greetings and welcome back people of Middle Earth the reason you're seeing this function on your screen is because I am a math nerd and I was bored and when I'm bored I like to play around with things so we have the logarithm of 1us e to the I Theta where I is of course the imaginary unit and we're going to play around with this thing to see what kind of results we can conjure up so I like integrals so I'm just going to integrate this thing from 0 to x with respect to Theta and solid I okay cool now the logarithm has a really cool series expansion we have log 1 - Z = to sum over K from 1 to Infinity of Z to the k / K so if I replace Z here by e to the I Theta we have log 1 - e to the I Theta equal to negative sum over K from 1 to Infinity of e to the i k Theta / K and the plan was to integrate this thing right so we integrate from 0 to x with respect to Theta of course terribly sorry about that so that gives us I here equal to negative of the sum over K from 1 to Infinity of 1 / K * the integral from 0 to X of e to the i k Theta D Theta which gives us us negative sum over K from 1 to Infinity of 1 K * e to the i k Theta divided I * K with the limits being 0 and x Now 1 / I is negative I so we have two negatives canceling out and we have I * the sum over the positive integers K of 1 / k^ 2 time e to the I KX minus E to the 0 being 1 so we have I * the sum over K from 1 to infinity and making use of the linearity of the summation operator we have e to the I KX / k^ 2 minus the sum over K from 1 to Infinity of 1 / k^2 now one of these series is quite familiar to us this guy over here is of course the zeta function evaluated at two which we know from the Bassel problem equals i^ 2/ 6 but I'm going to leave it written as Zeta 2 because that just looks really cool and for this guy over here we need the help of a special function in this case the poly logarithm of order s of a complex number Z this thing has a series expansion that is the sum over K from 1 to Infinity of Z to the k / K to the s for complex number Z having absolute value less than one however if s is a positive integer greater than one for example the case of the poly logi the D logorithm that is the poly log of order 2 we have e to the i x in place of Z which has an absolute value of one so this thing equals the sum over K from 1 to Infinity of e to the I KX / k^2 which is exactly what we have so that's quite nice and this implies that I here equals I * the DI logarith of e to the iix minus Zeta 2 okay cool that seems quite nice but of course we're not going to stop over here we'd like to derive some more cool stuff from our integral so how should we do that obviously we just attack it from a different methodology a more direct approach here would be to expand the complex exponential function using Oilers formula by which we have terribly sorry about that this thing equal to cosine Theta + I * sin Theta so I can be written as the integral from 0 to X of log 1 - cine Theta - I * sin Theta D Theta now for some Elementary trigonometry we know that this thing equals 2 sin 2 thet / 2 from one of the double angle formula for the cosine function and from the double angle formula for the S function sin Theta can be expanded in this manner as twice the product of s and cosine of theta / 2 Okay cool so I here equals the integral from 0 to X of log 2 sin Theta / 2 can be factored out and we're left with 1 more sin Theta / 2 - I * the cosine of theta / 2 D Theta now making use of the properties of the logarithm as well as the linearity of the integration operator we have integral 0 to X of log 2 sin Theta / 2 D theta plus the integral from 0 to X of I'll need a little bit of writing space log sin Theta / 2 - I * cine Theta / 2 D Theta and this thing here is actually another special function this is the Clausen function of this X term over here rather it's the negative of the Claus and function of X so this is negative cl2 of X Okay cool so we have negative chlorine molecule of X that was a horrible chemist dad joke and I will say nothing more and just move on to this term over here which is absolutely beautiful because again we can invoke ERS formula so sin Theta /2 we all know that this is is just cine of < / 2us Theta / 2 in Disguise and we have minus s of piun / 2us Theta / 2 as well and this thing equals e to the ne of I * < / 2us Theta / 2 and this is quite convenient because the logarithm and the exponential function will cancel out leaving behind only the argument up top so this means I equal Claus in function of xus I * the integral from 0 to X of < / 2 - Theta / 2 D Theta which is again quite trivial to evaluate we have piun / 2 Theta - th^ 2 over 4 with the limits being 0 and x so we have PK / 2x - x^2 / 4 and this implies that the Ral from 0 to X of log 1 - e to the I theta equals what exactly we have negative Claus in function of x - I * > / 2x - x^2 over 4 so we now have two strikingly different forms of exactly the same integral Clos forms for the for the same integral that is which is pretty cool because that means we can derive some stuff so equating these two and and expanding using 1 / I we have the D logarithm of e to the IX minus Zeta 2 equal to okay expanding by 1/ I means expanding by negative I so we have I * you know what I'm just writing this directly we have I * the Clausen function of X and of course wa negative I and I that's + i^ 2 so -1 and we havek / 2X X rather < / 2x + x^2 / 4 Okay cool so we have a pretty cool expansion for the D logarithm of e to the iix which is in terms of the Claus in function a polinomial in X and Zeta 2 so let me just write this out we have Zeta 2 plus I * Claus in function of xus I might as well expand this a bit so that would give me wait 2 piun x - x^2 / 4 correct so that means I can factor out the X leaving behind 2 piun - x which does look pretty dope and is there anything else we can derive well yeah why not we have the DI logarithm of e to the IX so expanding this in series form that is the sum from K = 1 to Infinity of e to the I KX / k^2 which again using O's formula this entire video is sponsored by Oiler by the way so this thing equals the sum over K from 1 to Infinity of cine of KX / k^ 2 plus I * the sum over K from 1 to Infinity of s of KX / k^2 so equating the real and imaginary parts we get a pretty cool form for the sum over K of cine kx/ k^ 2 which is in fact Zeta 2 terribly sorry about that and I might as well expand it in this at this stage anyway so we have pi^ S over 6 - x * 2 piun - x / 4 okay cool and equating the imaginary Parts gives us a cool looking yet obvious result in terms of the Claus and function so we have S of KX over k^2 sum over K from 1 to Infinity equal to am I missing any negative signs no way so we have clause in function of X I hope you enjoyed the video be sure to like And subscribe thank you see you next time