Transcript for:
Inverse Laplace Transform Lecture

[Music] welcome back okay so we have been deep into the wilderness of complex functions and complex variables derivatives and integrals koshi gorsa koshi integral formula and we are just about out of this wilderness there is one more lecture so i believe i have timed this so that this is the 13th lecture in this series on complex analysis this should cover about a two-week course and i've kind of saved one of the most interesting applications for last with all of the tools that we've learned throughout these two weeks throughout these 12 prior lectures we are now able to solve for things like the inverse laplace transform which involves a pretty gnarly integral in the complex plane okay so i'm not going to sugar coat it this is not going to be the easiest lecture you've ever seen in complex analysis but it's very very useful very very important and it's going to show you how how to solve for these super important inverse laplace transforms and i'm going to try to guide you through the high level steps that matter why we are making the choices we're making how these things work together but i'm also going to show you the nitty-gritty details okay so this is going to be a little bit like going to the gym or eating your broccoli it's going to make you stronger good thing we love our broccoli and our kale so i like to ask my students you know why do we care about the laplace transform in the first place maybe i'll maybe i'll just uh write down the inverse laplace transform and then we will talk about it a little bit so the uh the laplace transform takes a time domain function and maps it into a kind of generalized frequency domain function so it's kind of a generalization of the fourier transform for a certain class of problems and to get back from that frequency domain back to the time domain i need to take the inverse laplace transform which we call l inverse of some function f hat which is a function of that complex variable or frequency domain variable s my laplace variable and this is usually defined as 1 over 2 pi i it's a super hairy integral i'm going to write it out 1 over 2 pi i integral of gamma minus i infinity to gamma plus i infinity i'll tell you what this means in a minute of my frequency domain function f hat times e to the s t d s and this is the definition when i put that triangle up there you know that i've kind of just uh pulled this out of my hat this is the definition of the inverse laplace transform so again if i have a time domain function i laplace transform it to get a frequency domain function and to get back to the time domain maybe maybe i take my ordinary differential equation and i laplace transform it to make it easier to solve and i get my transfer function and my solution then i inverse laplace transform to get my solution as a function of time this is the inverse laplace transform function and specifically there are some conditions on this so this gamma here this is an integral in the complex plane you see this goes from minus i infinity to i infinity plus a gamma gamma is chosen gamma is chosen to be greater than the real part of all the poles f hat so poles are just things like one over s minus five there will be a pole at s equals five where it's not analytic that function's not analytic and i need gamma to be bigger than the real part of every pole of this function and i'll draw you a picture in a minute but this is a condition uh we also assume that t uh is strictly non-negative t is greater than or equal to zero so this works for for positive time uh into the future uh so i'm going to show you how to compute this integral using things like cauchy integral formula and the ml bound and things like that but just kind of philosophically i'll give you a little bit of a recap of why we care about the laplace transform we use the laplace transform all the time when we're solving ordinary and partial differential equations specifically you know partial differential equations that are parabolic like the heat equation where you know you can solve for them forward in time but not backward in time those are going to be very amenable to using the the laplace transform for solution we also use this all the time in differential equations to transform differential equations into algebraic equations that are easier to solve kind of with high school math and then you inverse laplace transform those to get the solutions back in time also very useful in control theory if you have input output systems you laplace transform both sides uh the outputs of the system and the inputs of the system and if you take that ratio that's the transfer function that tells you if i give a sinusoidal input what happens to the output of the system very useful for control systems and probably one of the most interesting and useful uh functions that we might want to inverse laplace transform so usually if you if you open up an engineering math book almost all of them have a big table in either the front or the back cover of all of the forward and inverse laplace transform pairs and oftentimes you just are told to memorize these and expected to memorize them and i mean that's okay you can remember what these things are but i want to show you how to actually derive from scratch what the inverse laplace transform is so one of the most inverse important inverse laplace transforms we care about so a specific example um example is the function f hat of s equals 1 over s minus a where a is often a real number a equals two or four or negative one okay um this is a laplace transform function a frequency domain function a transfer function that comes up all the time in differential equations this is in fact the uh transfer function for the simplest differential equation x dot equals a x okay and so for example if i laplace transform this uh differential equation i get you know the laplace transform of x dot equals uh s x technically it's s sorry this is x hat it's s x um you know minus my initial condition x at time zero but i'm gonna neglect that and so if i if i say the laplace transform of the left-hand side is s x the laplace transform of the right-hand side is a x hat and if i collect all of my x hats on one side then i get x hat equals so this would be s minus a times x hat equals i need that initial condition let me do this sorry minus x naught and that whole thing equals a times x hat this is actually the laplace transform of of x dot you should be catching me on this you should be yelling into your monitors so the laplace transform of the left-hand side is s the laplace variable times x-hat the laplace transform of x minus my initial condition and that equals the laplace transform of the right hand side which is a x hat and now if i collect all of my x hats i get s minus a x hat equals x naught and so i can rewrite that as x naught over s minus a so this transfer function one over s minus a comes up all the time it's this it's the the transfer function for the simplest differential equation with initial condition x not equals one okay that's that's what this transfer function is and i want to now that i've transformed my solution um so to find this transfer function now what i want to do is i want to inverse laplace transform to get x as a function of t the solution of my differential equation so i would inverse i inverse laplace transform kind of l negative 1 to get x of t this was in the laplace variable this is x hat as a function of s here i inverse laplace transform this now to get x as a function of t which again from differential equations we know that the solution is just x not e to the a t so we know that this is the answer and in principle that means that we know that the inverse laplace transform of 1 over s minus a is equal to e to the a t but i want to show you how to actually solve this inverse laplace transform using this integral formula in the complex plane because not all differential equations that we come up against are going to be as easy to solve like i just wrote down the solution of this differential equation off the top of my head we know the solution of linear differential equations uh x dot equals ax are e to the a t times the initial condition so i didn't technically need a laplace transform an inverse laplace transform but lots of times there will be uh differential equations that are hairy enough and gnarly enough that i really will want to use the laplace transform and inverse laplace transform to f to find the solution so that's what i'm going to do i'm going to derive today for you using everything you learned from complex function theory i'm going to derive that the inverse laplace transform of this transfer function this very important transfer function is e to the a t and we're going to do that using integrals in the complex plane that's what we're doing today good and i'm going to start by cooking up a really big gnarly contour integral for us to do this uh inverse laplace transform okay um so i'm trying to think of where i'm going to write this down i'm just going to erase this and then i'll probably fast forward okay good so i'm going to write down this specific inverse laplace transform integral for this specific transfer function that we're going to walk through today and then we're going to make an attack plan for how to actually solve this in the complex plane so the specific integral we're going to solve is the inverse laplace transform of 1 over s minus a and so i'm going to plug in 1 over s minus a for my function f hat that's my function f hat so this is going to equal 1 over 2 pi i integral of gamma minus i infinity to gamma plus i infinity of my function 1 over s minus a times e to the st e to the st over 1 over s minus a and i integrate with respect to the laplace variable ds so this is the integral i'm going to be trying to solve okay and what's kind of cool about this is that this looks right out of the gate like exactly the kind of thing i would want to use the cauchy integral formula for okay this is kind of a the poster child for solving an integral with the cauchy integral formula because i have an so s is a complex variable here s is kind of my z variable and so this is an analytic function in the numerator this is an analytic function divided by s minus a a kind of a transfer function with a simple pole at s equals a so this is exactly what we want to solve using uh the cauchy integral formula and so i'm going to write down a couple of a couple of i'm basically going to draw the contour integral that i want to solve and then we're going to make an attack plan for how to do this using everything i've told you about the complex plane so [Music] what i'm going to do is i'm going to draw in the complex plane here okay so this is the complex plane we have this pole at s equals a and so i'm just going to choose arbitrarily a place to drop down my pole we're going to say that this is where my pole a is i'm going to just for now say it's real valued it doesn't have to be real valued i'm just going to say it's real valued and positive so it's an unstable pole and i'm going to inverse laplace transform uh that system so one of the the kind of technical details of the laplace transform and you can go back to my video on the laplace transform to see is that when you do this integral you need this gamma to be to the right it has to have a real part that's bigger than the real part of this pole a in this transfer function so that line gamma so essentially this integral is along this big vertical contour integral that is offset from the from from the imaginary axis by gamma so this is kind of um you know gamma plus i omega or something like that and this is the integral that i want i want this big integral here that's what i've written down here and so the strategy that we're going to take is that we are going to essentially create a closed contour around this pole because we know how to solve integrals of closed contours of an analytic function divided by a simple pole so what i'm going to do is i'm going to connect like this i'm going to have a kind of contour i'm going to call this c plus and i'm going to have this contour i'm going to call c minus and then i'm going to draw a big circle so this is the classic integral that people use to solve this problem there's other ways of doing it but this is kind of the classic one because it's the easiest approach where now i would draw a big radius circle here cr okay and if i take the integral i want let's call this just uh c i the integral i want you know c one if i take c one plus c plus plus c r plus c minus this closed contour [Music] we know that we can compute that closed contour by the cauchy integral formula it's going to be a simple expression and if i took the limit as the radius goes to infinity if i blew this up so that the radius goes to infinity then this portion of that contour is the thing i want it goes from gamma minus i infinity to gamma plus i infinity and so so that that's how we're going to approach this problem let me annotate this a little bit this is literally um this here has a radius r okay this point here is gamma plus ir this point here is gamma minus ir and what we're going to do i'll just write out the integral we're going to solve we're going to say this is called a bromwich integral by the way a bromwich brom witch integral not to be confused with the brood witch integral the brood witch i don't know if any of you remember this the brood witch is one of the tastiest but most evil sandwiches known to humankind made with wheat harvested from hell's half acre mayonnaise made from the eggs of a deranged chicken this is why i actually wanted this to be the 13th lecture that's not the this is not the brood which integral this is the bromwich integral in texas there is a dairy chain called brahms that does make sandwiches and those witches are delicious this bromwich integral is this full large closed contour that i'm drawing this kind of backwards d we're going to call that c we're just going to call that c that's our contour c which is our bromwich integral and what we know because of cauchy integral formula is that we can solve the integral over this closed contour this big closed contour for any radius that encapsulates this for any radius at all actually for any radius this closed contour is going to include that simple pole at s equals a and i can use the cauchy integral formula to say that the integral of e to the s t over s minus a d s around that closed contour is equal to e to the a t now i'll just refresh what uh this is by let's let's write this down by the koshi integral formula which i believe uh was just a couple of lectures ago i showed you this kosher integral formula which says if you're integrating an analytic function in the complex plane divided by a simple pole s minus a or z minus a and i'm integrating it in a closed contour surrounding that pole i take the analytic function and i evaluate it at the pole and that's my answer and this is times 2 pi i i'm sorry i should have absolutely this is times 2 pi i that's what this contour integral equals and since i have uh divided by 2 pi i here those are going to cancel and this integral is going to equal e to the a t so let me just show you i've kind of held back the last piece of the logic of the puzzle here and i'm going to show you that right now which is the following this contour integral maybe i'm just going to do 1 over 2 pi i out here because this is 1 over 2 pi i and i'm just going to get rid of this here this contour integral which is easy to solve so this is easy easy easy to solve using coshy integral formula we can just write down the solution of this contour integral around the whole bromwich contour is e to the a t and we know that this is equal to the integral of the thing i want contour 1 plus the integral of c plus plus the integral of c minus plus the integral of c r okay so um this big bromwich integral is the thing i want this integral from gamma minus i infinity to gamma plus i infinity plus this little positive segment plus the big radius segment plus my little negative segment so it's four integrals that i've stitched into one mega brood which integral and what i'm going to show you is that um this equals you know so let's say in the in the limit as r goes to infinity c1 equals what we want and i'm going to prove that all three of these integrals add up to zero that in the limit as the radius goes to infinity as i scale this thing infinitely large so that this c1 becomes the integral i want this c plus and the c minus are going to be zero and the c r is also going to be zero so plus zero plus zero plus zero and what that shows kind of the logic here is that i can compute this big bromwich integral easily and it equals e to the a t and that in the limit as r goes to infinity is equal to my integral that i want which shows it proves that this integral that i have here is going to equal e to the a t okay so that's the punch line that is by far like that's the the headline here of this story is that if i can show that these three integrals equal zero and the limit as r goes to infinity we have then proven that this is the inverse laplace transform of uh one over s minus a which i kind of showed you from differential equations this is in fact the inverse laplace transform but we are going to to use complex integrals to show that okay so we all that we've used right now is we've cooked up a complex big gnarly contour the bromwich contour that i can break into four pieces and solving for each of these four pieces is going to be the hard part but once i've made this big contour integral for any radius r i can use koshi's integral formula very very easily to show that the the full bromwich integral equals e to the a t so if i can show that this segment this segment and this segment are zero then this full contour integral that was easy is just equal to my laplace transform my inverse laplace transform integral good that's the high level logic and now for the next i don't know 10 20 minutes i'm going to show you piece by piece that these three integrals are equal to zero which then will prove that this inverse laplace transform equals e to the a t and then i'll do a quick recap okay so that's what we're going to do now and i'm going to jump right in [Music] it turns out that these ones are kind of easy so i'm going to show those first we're going to use the ml bound to show that c plus and c minus go to zero in the limit as r goes to infinity uh okay so integral of uh c plus and integral of c minus we're going to use the m l bound which is kind of one of the easiest integral theorems in complex analysis is very trivial so the length of this integral path is always gamma length is always gamma regardless of the radius the radius could be 2 it could be 10 it could be a million all that's going to do is change the the the size of c1 and the size of this radius kind of scale it up and down but the length of c plus is always gamma and the length of c minus is always gamma so l is uh the l in the ml bound is always going to be gamma and so now what i have to do is i have to bound this portion of my integral and i have to show that this kind of gets smaller and smaller as my radius increases on this segment so i'm just going to do this for c plus so for c plus of e to the s t over s minus a d s what we have this is equal to the integral i'm actually going to write out what this integral equals it's the integral from gamma plus ir to just i r right we're just moving kind of in this real direction from gamma plus ir to ir uh again of e to the s t over s minus a d t nope d s sorry d s d s d s and so um essentially again i'm going to use some shorthand notation here to say that this is an integral uh from gamma to 0 of e to the x plus i r i'm just saying that my s variable my variable s equals uh x plus i r on this little segment where x varies from gamma to zero that's pretty pretty straightforward on this little segment my complex variable is x plus ir the imaginary part's always held constant at i r and x varies from gamma to zero so i'm integrating from gamma to zero of e to the x plus i r t divided by x plus i r minus a d x now not d s d dx and i'm going to check my notes because this could be something i would make a mistake on but this is correct and what we're going to show is that this is you know less than or equal to some m l bound so the length of this is gamma we're going to show that this quantity is equal to you know is is no greater than m on that whole path and we're going to show that that m gets smaller and smaller as r goes to infinity that's how ml bounds work and it's really really kind of elegant and simple and so m is the max of x between zero and gamma of e to the x plus i r t divided by x plus i r minus a kind of the magnitude of this this is again what we're doing for ml bounds and i'm going to claim that this quantity is always less than or equal to e to the gamma t over r now how do i want to do it um remember this we're going to use this a bunch in this lecture so i'm just going to write it down this is e to the real part plus imaginary part e to the x t plus i r t so if i have e to the x t plus i r t this is going to equal e to the x t times this is from euler's formula you know cos rt plus i sine rt and very very important property this has norm equal to one for any rt cosine something plus i sine something the magnitude the length of this vector is always equal to one so the length of e to the x t plus i r t is always equal to the the magnitude of the real part e to the x t okay this is the magnitude so the magnitude of this quantity here the biggest it can be is when x equals gamma so the maximum magnitude of this numerator is e to the gamma t pretty straightforward if that was a little fast just rewind and go through it again this is this is a cornerstone property of complex numbers that you know e to the u plus i v equals e to the u times cos v plus i sine v and this iv no matter what v is it can't contribute to the length of this vector um you know it doesn't contribute at all to the length of this vector v is essentially the angle and the the length of the vector or the radius is just e to the u the e to the real part okay and so the max value that this can uh the the length of this vector can take is e to the x when x is gamma okay so that's what we have here and similarly down here the largest length that this can have is let me see um x plus i r minus a okay so if i have um yeah the biggest this can be is if x equals a if x equals a then this is zero and that minus kind of gets cancelled out uh and this thing is going to be you know something which is on the the order of length of r let me uh is that 100 correct you know if x is between 0 and gamma and this is a fixed number then if i make r really really really really really really big then this quantity is going to dominate over those two and so this numerator is essentially going to be an r so what am i trying to say um yeah the length of this thing sorry i'm getting i'm mixing myself up let me this is important though for you to see me get confused because this is confusing um this numerator is always less than or equal to e to the gamma t good and in fact it its maximum value is exactly when x equals gamma and is exactly e to the gamma t so this should be for the numerator that's inequality that's the maximum value and it's divided by something and the max of this thing is attained when the new when the denominator is the smallest length it can possibly be and this thing has its smallest norm or length normally it would be r squared plus x minus a squared square root and so if x minus a is non-zero this thing is actually a really really big radius you know it's a bigger number it's a bigger length and so i'm saying that the maximum value is when this has its smallest norm this has its smallest value when x equals a and when x equals a which is in this interval then the norm of this is exactly equal to r okay so that's again that was a little bit confusing a little bit fast you might want to rewind that last two minutes if you if you think so so to find the max of this whole length i'm trying to find the value x where the numerator is as big as possible and where the denominator is as small as possible the biggest the numerator can get is e to the gamma t when x equals gamma the smallest the length of the denominator can get is r when x equals a so this is the ml bound m everywhere on this curve my function is has length no no larger than e to the gamma t over r and the nice thing about that is that now m and l so m is e to the gamma t over r l is a constant gamma so if i take the limit as r goes to infinity this thing goes to zero because i'm dividing by infinity i'm dividing a constant number times a constant number divided by infinity goes to zero that proves that c plus the plus part of my contour this integral equals zero and the c minus is exactly the same you don't i'm not going to like do a whole nother thing just for c minus it's exactly the same so both of these plus and minus portions of the integral equal zero in the limit as my radius goes to infinity as this thing gets bigger and bigger okay so those two are done and now i have to show that this big cr contour also goes to zero i'm going to erase my board and that's what we're going to do next okay so now i'm going to solve the hardest part of this okay so bear with me this is never going to be easy but this is important okay like we just spent two weeks learning all this interesting complex function theory and how to integrate in the complex plane and this is an actual legitimately useful and important application here is to solve the inverse laplace transform and so i really i really want to get to to show you so now i'm going to show that this big radius portion of the curve goes to zero as my radius goes to zero so i'm going to show that this equals zero this one's going to be a little bit trickier so ml bounds for integral of cr ml bounds don't work uh at least i've never been able to get it to work don't work and so we're going to use some tricks these are tricks that i must have picked up somewhere in my education i don't remember where i got these uh you will someday forget where you learned them but the basic tricks is we're going to try polar coordinates and that's why we extended this instead of just having a big radius coming directly from uh c1 that's why we had c plus and c minus to move these over to a place where we could actually use polar coordinates so this point is going to be r you know e to the i theta so there's going to be an r cosine theta and an r sine theta so we're going to say x equals minus r cosine theta and we're going to say y equals our sine theta and we're going to do an integral basically from theta equals minus pi halves to pi halves something like that okay we'll we'll see and you'll have to check that i'm doing the right bounds of integration uh and so essentially what we're doing is you know we're solving for this portion of the bromwich integral here where you know theta is going to start at and i'm really abusing abusing polar notation here this is i'm calling this minus pi halves because i have a minus r cosine theta and i have you know this goes to theta equals plus pi halves and it's going counterclockwise i am pretty sure there's a better way of integrating from plus pi halves to minus pi halves with a plus r cosine theta and maybe i should do that but i'm just going to stick with my notes and maybe as an exercise you can try to you know use the normal polar convention of x equals plus r cosine theta and change this to theta equals plus pi halves to minus pi halves i think you know we're pretty good at uh changing the definition of where things start and what's positive and negative so i think this will be okay and so we're again trying to to kind of show what the properties of these two terms in this integral are uh across this contour from theta equals minus pi halves to pi halves so i'm going to isolate this and break this up i have the uh maybe i'll switch colors to make it a little more interesting so i have my e to the s t portion i'm going to look at the norm of that again the length the norm uh if i have a if i haven't you know my s t has a real part let's call that you know x plus i y the norm of uh you know e to the s t the norm of that is going to equal e to the x it's going to be the length of e to the real part so the norm of this is going to be minus r e to the minus r cosine theta that's that's going to be the norm of this thing so that's uh e to the minus r cos theta t that's the norm of of the numerator is this thing here uh and the norm of s minus a the denominator here is similarly going to equal the norm of and i'm just going to plug everything in here um i think it's minus r cos theta um plus i r sine theta minus a and i put the minus a here because i'm going to group these real valued parts and this complex value part separately but but the s is the r cosine and r sine and the a is right here okay so that's the norm and that norm is equal to uh this is kind of a mess but it's the square root of r cos theta plus a quantity squared plus r sine theta quantity squared the square root of that it's literally the the norm of this is the real part squared plus the imaginary part squared square root that's this and if i do a little fancy arithmetic here and i cancel out some stuff i'm going to get an r cos squared and an r sine squared those are going to add up to an r squared so i'm going to get an r squared i'm going to get a plus 2 r cos theta plus a squared okay that's what all of this simplifies to you can verify that and i'm going to claim that this is always greater than or equal to r minus a to the norm of r minus a okay um why is that true so this is kind of [Music] um well the norm of r minus a is r squared plus a squared minus 2 a r and this is i'm sorry i missed an a when i expanded this out this should be 2 r times a times cosine theta and yeah so the norm of r minus a is just r squared plus a squared minus 2ar so it looks almost identical to this except that up here this 2 ar is multiplied by cos theta and cos theta can never be bigger than 1 so this expression is always strictly less than or equal yeah th this expression here is always um i this should be a greater than or equal i'm sorry this should be less than or equal this expression here the length of this r squared plus a squared i'm sorry i'm like uh it's been a real wilderness of complex variables uh this was correct i want you to i want you to really see why this is true so i'm gonna derive this uh i'm gonna i'm gonna show this again this is this is the hardest integral so it makes sense that we're getting a little confused here we're going to do something kind of like the ml bound we're finding the norm of this and the norm of this and i'm going to use some argument to show that as r goes to a large number you know the ratio goes to zero something like that and so i need an expression for the norm of s minus a and i've done some math and i've done a trick and i've done another trick and now i am showing that this expression here is always bigger than the norm of r minus a the norm of r minus a is this quantity here and this is minus 2 a r so this thing is always you know detracting from the norm it's making it smaller [Music] and unless cosine of theta is exactly equal to -1 this term is detracting less or maybe even adding to the norm of r minus a so so i want you to just think about why this is always bigger than or equal to this because of the properties of cosine so if cosine is negative one then this equals this but if cosine is anything you know bigger than negative one you know it goes between negative one and plus one this expression is greater than this norm here and we're going to use this property kind of like we use the ml bound good okay so uh that is very useful and so what this means is i'm gonna say so the integral of the integral over cr of e to the st over s minus a d s is um let's say that the the norm of this whole thing is strictly less than this integral we're going to now integrate over thetas because we've gone to polar coordinates the um integral of of the with respect to theta of this thing e to the minus r cos theta t divided by uh norm of r minus a times and we're integrating from minus pi over 2 to pi over 2 d theta and [Music] last important thing when we're integrating in polar coordinates this is r d theta because we're at a radius r and so we're going r times d theta is uh you know every little segment here d s is r d theta because we're at a big radius r okay this is radius r and uh so now what we're going to try to do is we're going to try to show that this thing goes to zero as r goes to infinity okay um and we can also even simplify this more we can this is an even function this is an uh an even function and so we can say that this equals uh 2r over r minus a norm integral from 0 to 2pi sorry pi pardon my french pi over 2 laplace was french of e to the minus r t cos theta d theta we're integrating with respect to theta we pulled out the r's this is a really ugly pi over 2. and i was able to do this i was able to do this integral to say that this this integral from minus pi over 2 to pi over 2 equals this because it's an even function so i picked up this value of 2 it's twice this half of that even function okay we're getting really close thank you for bearing with me just a little bit farther if this is just boring you to tears you can skip to the next chapter mark but i would stay with me because we're almost there and so there are a couple of other tricks we're going to do here um basically we're just going to keep doing tricks until we can force this thing into a reasonable reasonable solution so the last trick is that cos theta is always greater than or equal to 1 minus 2 theta over pi on the interval theta between zero and pi halves okay on that interval and so e to the minus r cos theta t is always less than or equal to e to the minus r t times this value here this 1 minus 2 theta over pi good and i could make my minus go in here and say that that's of course equal to e to the r t times 2 pi over theta 2 theta over pi minus 1. okay so i've just used uh kind of every trick at my disposal every trigonometric trick i have you can again verify that this is true this is just um like kind of a basic trigonometric fact that you should uh verify if you don't believe me and so finally this integral here this uh integral here we're going to say is um this thing here is less than or equal to 2 r over norm of r minus a of this integral which is 2 r over norm of r minus a of this integral from 0 to pi halves of now this expression here e to the rt theta 2 theta over pi minus 1 d theta it's actually strictly less than or equal to this and this is something we can actually solve this is now an integral e to the theta constant plus a constant d theta we can actually solve this closed form so we're basically there and that equals 2 r over norm of r minus a pi over 2 r t e to the r t 2 theta over pi minus 1 this whole thing evaluated at the bounds of integration pi over 2 and 0. and if you plug all of this stuff in here so at this point this just became a standard integral you do know how to solve this is the answer we're going to evaluate it and that just equals pi over norm of r minus a t one minus e to the minus r t and e to the r t e to the minus r t as r goes to infinity this goes to zero so this whole expression goes to one as r goes to infinity and pi over norm of r minus a the norm of r goes to infinity as r goes to infinity so this whole thing goes to zero as r goes to infinity whoa that was a mess okay this is a nightmare that's why i think of this as the broodwich inner rule because this is the most evil integral uh you'd use so many tricks to get this uh this big radius cr to go to zero as r goes to infinity but it is technically possible right so we used a bunch we switched to polar coordinates we derived these expressions for what these numerator and denominator are bigger than or smaller than and then we did more trig substitutions but eventually we showed and you'll have to check my steps like i require you know i expect my students to actually check these steps that i did in the middle just to really convince yourself and kind of go back and see why i was doing it in the first place okay good so that was the tough part now i am going to summarize everything so the big summary here is that i'm trying to solve this inverse laplace transform integral because that's super duper useful in differential equations both ordinary and partial in control theory and in tons of other fields of engineering is to be able to laplace transform and inverse laplace transform and specifically if i want to inverse laplace transform something like this function one over s minus a which comes up all the time in linear differential equations i essentially need to solve for this very very large vertical integral in the complex plane the integral from gamma minus i infinity to gamma plus i infinity and so what i do instead is i create a large closed contour called the bromwich integral that has the integral i want c1 plus three other segments that i don't want but i close this loop around that pole so that i can use koshi's integral formula because i can solve i can integrate this closed contour integral around c of my function i'm trying to inverse laplace transform easily using cauchy integral formula it's essentially trivial because this is an analytic function and there's a it's divided by a simple pole at s equals a then the closed contour integral of everything equals e to the a t and that's the sum of the integral i want plus this this little segment plus this little minus segment plus this big radius contour here and so if i can show that in the limit as r goes to infinity that these three terms go to zero and in the limit as r goes to infinity this segment becomes the integral i want so if i can show that these three go to zero then the integral i want equals e to the a t and we showed it took i guess 25 30 minutes that these three integrals equal zero in the limit as r goes to infinity so therefore we have shown that the integral we want the laplace the inverse laplace transform by koshi integral formula and some heinous math is equal to this e to the a t so that's super duper cool and i think now you are armed with some of the most powerful tools in mathematics you know how to integrate in the complex plane you know how to differentiate in the complex plane and now you can solve these are some of the nastiest integrals out there if you can solve these you can solve anything all right i'm really proud of you all for sticking with me through you know 50 minutes of complex integrals bromwich integrals this is the real deal i'll see you next time thank you