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Perhaps some may enjoy giving this one a go. It may be kind of challenging.

$$\int_{0}^{\infty}\frac{x}{x^{2}+b^{2}}\log\left(\frac{x^{2}+2ax\cos(t)+a^{2}}{x^{2}-2ax\cos(t)+a^{2}}\right)dx$$

$$=\frac{\pi^{2}}{2}-\pi t+\pi\tan^{-1}\left(\frac{(a^{2}-b^{2})\cos(t)}{(a^{2}+b^{2})\sin(t)+2ab}\right), \;\ a,b>0; \;\ 0<t<\pi$$

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  • $\begingroup$ I have a rough idea. Convert to complex exponential then differentiate with respect to t. Then we can solve the resultant integral using residues. It remains to find the anti derivative of the result. $\endgroup$ – Zaid Alyafeai Jun 8 '14 at 18:28
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$$I(t)=\int_0^{\infty} \frac{x}{x^2+b^2}\ln\left(\frac{x^2+2ax\cos(t)+a^2}{x^2-2ax\cos(t)+a^2}\right)\,dx$$ $$I'(t)=\int_0^{\infty} \frac{x}{x^2+b^2}\left(\frac{-2ax\sin(t)}{x^2+2ax\cos(t)+a^2}-\frac{2ax\sin(t)}{x^2-2ax\cos(t)+a^2}\right)\,dx$$ $$\begin{aligned} \Rightarrow I'(t) \,& =-2a\int_0^{\infty}\frac{x}{x^2+b^2}\left(\frac{x\sin(t)}{x^2+2ax\cos(t)+a^2}+\frac{x\sin(t)}{x^2-2ax\cos(t)+a^2}\right)\,dx\\ & =-2a\,\Im\left(\int_0^{\infty} \frac{x}{x^2+b^2}\left(\frac{e^{it}}{x+ae^{it}}+\frac{e^{it}}{x-ae^{it}}\right)\,dx\right)\,\,\,\,\,\,\,\,(1)\\ & = -2a\,\Im\left(2e^{it} \int_0^{\infty} \frac{x^2}{(x^2+b^2)(x^2-a^2e^{2it})}\,dx \right)\\ & = -2a\,\Im\left(2e^{it}\frac{\pi}{2(b-iae^{it})}\right)\,\,\,\,\,\,\,\,(2)\\ & = -2\pi a\,\Im\left(\frac{e^{it}}{b-iae^{it}}\right)\\ & = -2\pi a\,\frac{b\sin t+a}{b^2+2ab\sin t+a^2}\,\,\,\,\,\,(3)\\ & = -\pi\left(1+\frac{a^2-b^2}{b^2+2ab\sin(t)+a^2}\right)\\ \end{aligned}$$

$$\Rightarrow I(t)=-\pi t-2\pi\arctan\left(\frac{(a^2+b^2)\tan(t/2)+2ab}{a^2-b^2}\right)+C\,\,\,\,\,\,(4)$$ Since $I(\pi/2)=0$, $C=\frac{\pi^2}{2}+2\pi\arctan\left(\frac{a+b}{a-b}\right)$. $$\Rightarrow I(t)=\frac{\pi^2}{2}-\pi t-2\pi\arctan\left(\frac{(a^2+b^2)\tan(t/2)+2ab}{a^2-b^2}\right)+2\pi\arctan\left(\frac{a+b}{a-b}\right)$$ The above expression can be shown equivalent to as shown in OP by simplifying arctangents: $$\begin{aligned} -2\pi\left(\arctan\left(\frac{(a^2+b^2)\tan(t/2)+2ab}{a^2-b^2}\right)-\arctan\left(\frac{a+b}{a-b}\right)\right) &=2\pi\arctan\left(\frac{a-b}{a+b}\frac{1-\sin(t)}{\cos t}\right)\\ &=\pi\arctan\left(\frac{(a^2-b^2)\cos(t)}{(a^2+b^2)\sin(t)+2ab}\right) \end{aligned}$$ Hence $$I(t)=\frac{\pi^2}{2}-\pi t+\pi\arctan\left(\frac{(a^2-b^2)\cos(t)}{(a^2+b^2)\sin(t)+2ab}\right)$$


Proof of $(1)$: $$\begin{aligned} \frac{x\sin t}{x^2+2ax\cos(t)+a^2}&=\frac{1}{2i}\frac{x(e^{it}-e^{-it})}{x^2+axe^{it}+axe^{-it}+a^2}\\ &= \frac{1}{2i}\left(\frac{e^{it}}{x+ae^{it}}-\frac{e^{-it}}{x+ae^{-it}}\right)\\ &= \Im\left(\frac{e^{it}}{x+ae^{it}}\right)\,\,\,\,\,\,\left(\because \Im(z)=\frac{1}{2i}(z-\bar{z})\right)\\ \end{aligned}$$

Similarly it can be shown that: $$\frac{x\sin t}{x^2-2ax\cos(t)+a^2}=\Im\left(\frac{e^{it}}{x-ae^{it}}\right)$$


Proof for $(2)$: $$\begin{aligned} \int_0^{\infty} \frac{x^2}{(x^2+b^2)(x^2-a^2e^{2it})}\,dx &=\frac{1}{b^2+a^2e^{2it}}\int_0^{\infty} \left(\frac{x^2}{x^2+b^2}-\frac{x^2}{x^2-a^2e^{2it}}\right)\,dx\\ &= -\frac{1}{b^2+a^2e^{2it}}\int_0^{\infty} \left(\frac{b^2}{x^2+b^2}+\frac{a^2e^{2it}}{x^2-a^2e^{2it}}\right)\,dt\\ &=\frac{1}{b^2+a^2e^{2it}} \left(\frac{\pi b}{2}+\frac{\pi iae^{it}}{2}\right)\\ &=\frac{\pi}{2}\frac{1}{(b+iae^{it})(b-ia^{it})}(b+iae^{it})\\ &=\frac{\pi}{2(b-iae^{it})}\\ \end{aligned} $$


Proof for $(3)$:

$$\begin{aligned} \Im\left(\frac{e^{it}}{b-iae^{it}}\right) &= \Im\left(\frac{\cos(t)+i\sin(t)}{(b+a\sin(t))-ia\cos(t)}\right)\\ &=\Im\left(\frac{(cos(t)+i\sin(t))(b+a\sin(t)+ia\cos(t))}{(b+\sin(t))^2+a^2\cos^2(t)}\right)\\ &= \frac{b\sin (t)+a}{b^2+2ab\sin(t)+a^2}\\ \end{aligned}$$


Proof for $(4)$: $$\begin{aligned} \int \frac{a^2-b^2}{b^2+2ab\sin(t)+a^2}\,dt &=(a^2-b^2)\int \frac{\sec^2(t/2)}{(a^2+b^2)(1+\tan^2(t/2))+4ab\tan(t/2)}\,dt\\ &= 2(a^2-b^2)\int \frac{dp}{(a^2+b^2)(1+p^2)+4abp}\,\,\,\,\,\,\,\,\,\left(p=\tan(t/2)\right)\\ &=2\frac{a^2-b^2}{a^2+b^2}\int\frac{dp}{p^2+\frac{4abp}{a^2+b^2}+1}\\ &=2\frac{a^2-b^2}{a^2+b^2}\int \frac{dp}{\left(p+\frac{2ab}{a^2+b^2}\right)^2+\left(\frac{a^2-b^2}{a^2+b^2}\right)^2}\\ &=2\arctan\left(\frac{(a^2+b^2)\tan(t/2)+2ab}{a^2-b^2}\right)+C\\ \end{aligned}$$

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  • $\begingroup$ Nice answer. +1 $\endgroup$ – Random Variable Jun 8 '14 at 19:45
  • $\begingroup$ @RandomVariable: Thanks a lot! :) Do you have some ideas about showing that my last expression is same as shown in OP? I am really struggling with that. $\endgroup$ – Pranav Arora Jun 8 '14 at 19:53
  • $\begingroup$ @RandomVariable: Sorry, I was being hasty, I have proved that the expression is equivalent to shown in OP. :) $\endgroup$ – Pranav Arora Jun 8 '14 at 21:10
  • $\begingroup$ Nice work PA. :) very thorough. $\endgroup$ – Cody Jun 8 '14 at 22:04
  • $\begingroup$ @Cody: Thank you! Glad to know you like it. :) $\endgroup$ – Pranav Arora Jun 9 '14 at 1:53

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