A tricky arctan integral

Hey there folks, this is your boy Kamal once again and today we have yet another fascinating integral. We have the integral from zero to infinity of arc tangent of tangent square x over x squared dx. And I've actually sourced this integral from a really cool website called dailyintegral.com which as the name implies gives you integrals every day to solve and those integrals are pretty damn cool. I highly recommend you check that website out whoever is maintaining that and managing that. Absolutely brilliant stuff, brilliant job. Anyway so the objective is to solve this integral which by the way was actually represented as an integral from negative infinity to infinity but that of course makes sense because the integrand is an even function of x so I'm just going to solve in the integral from zero to infinity. You can double the result at the end to get the result of the actual integral from the website. Now the problem here is that we have the arc tangent function of tangent square x which is not exactly very hospitable is it? So we could try to get rid of it and by getting rid of it I mean we first define the integral function i of a parameter alpha as the integral from zero to infinity of arc tangent of alpha times tangent square x over x squared dx. And you guessed it we'll try to differentiate this thing because the derivative of arc tangent is quite nice to work with. So we have i prime of alpha on the left and on the right we have the integral from zero to infinity of the partial derivative with respect to alpha on switching up the order of the derivative and integral operators of course of arc tangent of alpha times tangent square x over x squared dx. Now because we're differentiating partially that means we can just ignore the one over x term one over x squared term for differentiating purposes and differentiating the arc tangent yields one plus alpha squared times tangent to the fourth power of x and because of the chain rule we do have this tangent square x term up top. Okay cool that actually looks like we've made things worse but they do look cooler I mean this integral is a lot cooler or slightly cooler than the integral we started off with and I'll take that as a good sign. And now we do have a tangent square up top so we might as well expand that as sine square over cosine square so that we have sine square x over x squared and I think you guys have figured out where this is going over cosine square of x times one plus alpha squared times tangent to the fourth power of x and of course now we have an opportunity to apply Lobachevsky's integral formula which I have made a couple videos about. I don't think I formally proved this variation the one with a sine square x over x squared term but I should probably make a video on that. I will link a related video in the description box. Anyway so if you have the following structure that is the integral from zero to infinity of sine square x over x squared times f of x dx where by the way f of x is even so f of x here equals f of negative x and we have f of x plus or minus pi equal to f of x in other words f is pi periodic. In that case the integral sorts out to the integral from zero to pi over two of simply f of x dx and we notice that our function here that is cosine square x that is even tangent to the fourth power of x again even so this thing is clearly even and this is also pi periodic because tangent itself is pi periodic and although cosine of x plus pi would be negative cosine x if I recall correctly the square is still positive so we're all good. This thing satisfies both criteria for I'll call this criteria a and criteria b it satisfies both criteria a and criteria b for a Planck-Lobachevsky's formula. So all of this implies that the derivative of I with respect to alpha is now just the integral from zero to pi over two of dx over cosine square x times one plus alpha squared times tangent to the fourth power of x and that looks slightly better I mean it still looks quite hostile if you ask me but perhaps a transformation or two could do the trick. So what if we let tangent x equal u which would imply that secant square x dx is equal to du and of course secant square is just one over cosine square so those are the ingredients for our differential element and we have them and as x approaches zero terribly sorry about that we have u approaching zero and as x approaches pi over two we have u approaching infinity of course. Okay cool of course x is supposed to approach pi over two from the left which it is anyway so this implies that I prime of alpha is now the integral from zero to infinity of du over one plus alpha squared times u to the fourth power and maybe I should have just clubbed this into one transformation but it's cool I could let root alpha times t times u equal t which implies that du here terribly sorry about my cursive du equals dt over root alpha and that still does not alter the limits of integration so we have one over root alpha terribly sorry about that times the integral from zero to infinity of dt over one plus t to the fourth power which again looks like quite an intimidating integral but we know complex analysis so we know about the beta function and a result connected to the beta function that I have derived a while back that is the integral from zero to infinity of x to the s minus one dx over one plus x to the k if I recall correctly this converges to pi over k times the cosecant of pi s over k now s here is clearly equal to one and k here is equal to four so that means I prime of alpha terribly sorry about that I prime of alpha is one over root alpha times pi over four times the cosecant of yet again pi over four and cosecant of pi over four would be root two if I'm not wrong so that means we have pi over four times root two over alpha okay cool so that's I prime completely in terms of the alpha parameter and we should now recover back the integral function by integrating with respect to the alpha parameter so that we have the integral pi over four times the integral rather wait we also have pi root two over four which is two times two so this is just pi over two root two times the integral of alpha to the minus one half d alpha which of course is pretty elementary to evaluate we have pi over two times root two alpha to the one half over one half and this should cancel with the one half outside resulting in a pi times root alpha over two plus a constant of integration c and on the left of course we have I of alpha this constant is pretty easy to evaluate as well since alpha equal to zero implies that I of alpha is equal to zero or I could just write this in one line I of zero is equal to zero implies that zero is equal to pi times zero plus c in other words the constant of integration is indeed a big fat zero again quite convenient so this implies that I of alpha my friends is pi times root alpha over two we plug in alpha equal to one and this gives us the integral from zero to infinity of arc tangent of tangent square x terribly sorry about that over x squared equal to a very beautiful looking result there that is pi times one over root two or pi over root two and of course the integral from negative infinity to infinity of arc tangent of terribly sorry about that tangent square x over x squared something's wrong with the exponents and this whole thing equals two times that result so two over root two is root two so pi times root two is the result of the other target integral I hope you enjoyed the video be sure to like and subscribe drop me a follow on instagram as well and I will see you guys in the next one thank you