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Hi I am trying to prove this $$ I:=\int_0^{\pi/4}\log\left(\tan\left(x\right)\right)\, \frac{\cos\left(2x\right)}{1+\alpha^{2}\sin^{2}\left(2x\right)}\,{\rm d}x =-\,\frac{\pi}{4\alpha}\,\text{arcsinh}\left(\alpha\right),\qquad \alpha^2<1. $$ What an amazing result this is! I tried to write $$ I=\int_0^{\pi/4} \log \sin x\frac{\cos 2x}{1+\alpha^2\sin^2 2x}-\int_0^{\pi/4}\log \cos x \frac{\cos 2x}{1+\alpha^2\sin^2 2x}dx $$ and played around enough here to realize it probably isn't the best idea. Now back to the original integral I, we can possibly change variables $y=\tan x$ and re-writing the original integral to obtain $$ \int_0^{\pi/4}\log \tan x \frac{\cos 2x}{1+{\alpha^2}\big(1-\cos^2 (2x)\big)}dx=\int_0^1 \log y \frac{1-y^2}{1+y^2}\frac{1}{1+{\alpha^2}\big(1-(\frac{1-y^2}{1+y^2})^2\big)}\frac{dy}{1+y^2}. $$ Simplifying this we have $$ I=\int_0^1\log y \frac{1-y^2}{1+y^2}\frac{(1+y^2)^2}{(1+y^2)^2+4\alpha^2y^2}\frac{dy}{1+y^2}=\int_0^1\log y \frac{1-y^2}{(1+y^2)^2+4\alpha^2y^2}dy $$ Another change of variables $y=e^{-t}$ and we have $$ I=-\int_0^\infty \frac{t(1-e^{-2t})}{(1+e^{-2t})^2+4\alpha^2 e^{-2t}} e^{-t}dt $$ but this is where I am stuck...How can we calculate I? Thanks.

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  • $\begingroup$ One thing that comes into my mind is integral f(x) = integral f(a-x) on interval (0,a) Perhaps worth looking into it? Two times pi/4 is pi/2 and so sines and cosines switch...just a thought $\endgroup$ – imranfat May 16 '14 at 18:09
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Integrate by parts; then you get that

$$I(\alpha) = \left [\frac1{2 \alpha} \arctan{(\alpha \sin{2 x})} \log{(\tan{x})} \right ]_0^{\pi/4} - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\alpha \sin{2 x}}$$

The first term on the RHS is zero. To evaluate the integral, expand the arctan into a Taylor series and get

$$I(\alpha) = -\frac12 \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \alpha^{2 k} \int_0^{\pi/2} du \, \sin^{2 k}{u} = -\frac{\pi}{4} \sum_{k=0}^{\infty} \frac{(-1)^k}{2 k+1} \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k}$$

A little manipulation leads us to

$$\alpha I'(\alpha) +I(\alpha) = -\frac{\pi}{4} \sum_{k=0}^{\infty} (-1)^k \binom{2 k}{k} \left (\frac{\alpha}{2} \right )^{2 k} = -\frac{\pi}{4} \frac1{\sqrt{1+\alpha^2}}$$

The LHS is just $[\alpha I(\alpha)]'$, so the solution is

$$I(\alpha) = -\frac{\pi}{4} \frac{\operatorname{arcsinh}(\alpha)}{\alpha} $$

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My first step is similar to Ron Gordon but then I took a different route. From integration by parts, the given integral can be written as: $$-\frac{1}{a}\int_0^{\pi/4} \frac{\arctan(\alpha \sin(2x))}{\sin (2x)}\,dx$$ Consider $$I(a)=\int_0^{\pi/4} \frac{\arctan(a \sin(2x))}{\sin (2x)}\,dx$$ Differentiate both the sides wrt $a$ to obtain: $$I'(a)=\int_0^{\pi/4} \frac{1}{1+a^2\sin^2(2x)}\,dx$$ Use the substitution $a\sin(2x)=t$ to obtain: $$I'(a)=\frac{1}{2}\int_0^a \frac{dt}{\sqrt{a^2-t^2}(1+t^2)}$$ Next use the substitution $t=a/y$ to get: $$I'(a)=\frac{1}{2}\int_1^{\infty} \frac{y}{\sqrt{y^2-1}(a^2+t^2)}\,dy$$ With yet another substitution which is $y^2-1=u^2$, $$I'(a)=\frac{1}{2}\int_0^{\infty} \frac{du}{u^2+a^2+1}$$ The final integral is trivial, hence: $$I'(a)=\frac{\pi}{4\sqrt{a^2+1}}$$ Integrate both sides wrt $a$ to get: $$I(a)=\frac{\pi}{4}\sinh^{-1}a+C$$ It is easy to see that $C=0$, hence with $a=\alpha$, $$-\frac{1}{a}\int_0^{\pi/4} \frac{\arctan(\alpha \sin(2x))}{\sin (2x)}\,dx=-\frac{\pi}{4\alpha}\sinh^{-1}\alpha$$ $\blacksquare$

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I use on Ron's results from integrating by parts:

$$\alpha I(\alpha) = - \int_0^{\pi/4} dx \frac{\arctan{(\alpha \sin{2 x})}}{\sin{2 x}},$$

As an alternative way to complete the problem, use the method differentiating under the integral sign:

$$\frac{d}{d\alpha}(\alpha I(\alpha)) = -\int_{0}^{\pi/4}\frac{dx}{\alpha^2\sin^2{2x}+1}=-\frac{\pi}{4\sqrt{1+\alpha^2}}\\ \implies \alpha I(\alpha) = -\frac{\pi}{4}\int_{0}^{\alpha}\frac{d\tilde\alpha}{\sqrt{1+\tilde\alpha^2}} = -\frac{\pi}{4}\sinh^{-1}{\alpha}.$$

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  • $\begingroup$ Ah, you beat me to it, I used the same method but you posted it while I was writing the solution. :) $\endgroup$ – Pranav Arora May 16 '14 at 19:27
  • $\begingroup$ @PranavArora Well no wonder yours took more time to write. Just look at all the unnecessary gymnastics you decided to throw in! ;) P.S. Thank you very much for showing me that there is a latex command for Little Box. One time out of curiosity I tried \littlebox to see if it did anything but to no avail. $\endgroup$ – David H May 16 '14 at 19:38

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