When I showed to my brother how I proved
\begin{equation}
\int_{0}^{\!\Large \frac{\pi}{2}} \ln \left(x^{2} + \ln^2\cos x\right) \, \mathrm{d}x=\pi\ln\ln2
\end{equation}
using the following theorem by Mr. Olivier Oloa
\begin{equation}{\large\int_{0}^{\!\Large \frac{\pi}{2}}} \frac{\cos \left( s \arctan \left(-\frac{x}{\ln \cos x}\right)\right)}{(x^2+\ln^2\! \cos x)^{\Large\frac{s}{2}}}\, \mathrm{d}x = \frac{\pi}{2}\frac{1}{\ln^{\Large s}\!2}\qquad,\;\text{for }-1<s<1.\end{equation}
He showed me the following interesting formula
\begin{equation}
\int_{0}^{\!\Large \frac{\pi}{2}} x\csc^2(x)\arctan \left(\alpha \tan x\right)\, \mathrm{d}x
=\frac{\pi}{2}\,
\ln\left(\left[1 + \alpha\right]^{1 + \alpha}
\over \alpha^\alpha\right)\,,\qquad
\mbox{for}\ \alpha > 0\tag{✪}.
\end{equation}
I tried several values of $\alpha$ to check its validity (since he always messes around with me) and the numerical results match the output of Mathematica $9$. The problem is how to prove this formula since he didn't tell me (as always). I tried Feynman's integration trick and I arrived to the following result:
\begin{equation}\partial_\alpha\int_{0}^{\!\Large \frac{\pi}{2}} x\csc^2(x)\arctan \left(\alpha \tan x\right)\, \mathrm{d}x = \int_{0}^{\!\Large \frac{\pi}{2}} \frac{x\cot x}{\cos^2x+\alpha^2 \sin^2 x}\, \mathrm{d}x\end{equation}
but I am having difficulty to crack the very last integral. Could anyone here please help me to prove the formula $(✪)$ preferably with elementary ways (high school methods)? Any help would be greatly appreciated. Thank you.
Best Answer
This approach is similar to Ron Gordon's answer. We introduce a new variable $\beta$ before differentiating.
By substitution $\tan x=u$, $$\int_{0}^{\!\Large \frac{\pi}{2}} x\csc^2(x)\arctan \left(\! \alpha \tan x\right)\, \mathrm{d}x=\int_{0}^{\infty}\frac{\tan^{-1}u\tan^{-1}\alpha u}{u^2}\,\mathrm{d}u$$
We will prove $$\int_0^{\infty}\frac{\tan^{-1}\alpha u\tan^{-1}\beta u}{u^2}\,\mathrm{d}u=\frac\pi2\log\left(\frac{(\alpha+\beta)^{\alpha+\beta}}{\alpha^\alpha\beta^\beta}\right)\tag{1}$$
Differentiation gives \begin{align}\partial_\alpha\partial_\beta\int_0^{\infty}\frac{\tan^{-1}\alpha u\tan^{-1}\beta u}{u^2}\,\mathrm{d}u&=\int_{0}^{\infty}\frac{\mathrm{d}u}{(1+\alpha^2 u^2)(1+\beta^2 u^2)}\\&=\frac1{\alpha^2-\beta^2}\int_{0}^{\infty}\frac{\alpha^2}{1+\alpha^2u^2}-\frac{\beta^2}{1+\beta^2u^2}\,\mathrm{d}u\\&=\frac1{\alpha^2-\beta^2}\int_{0}^{\infty}\frac{\alpha}{1+u^2}-\frac{\beta}{1+u^2}\,\mathrm{d}u\\&=\frac\pi{2(\alpha+\beta)}\end{align} for $\alpha\ne\beta$. The case $\alpha=\beta$ can be proved by letting $\alpha\to\beta$. Then $$\partial_\beta\int_0^{\infty}\frac{\tan^{-1}\alpha u\tan^{-1}\beta u}{u^2}\,\mathrm{d}u=\frac{\pi}{2}(\log(\alpha+\beta)-C_1(\beta))$$ If $\alpha\to0$ we have $0=\frac\pi2(\log\beta-C_1(\beta))$ so $C_1(\beta)=\log\beta$. Integrating by $\beta$,$$\int_0^{\infty}\frac{\tan^{-1}\alpha u\tan^{-1}\beta u}{u^2}\,\mathrm{d}u=\frac{\pi}{2}((\alpha+\beta)\log(\alpha+\beta)-\beta\log\beta-C_2(\alpha))$$ Let $\beta\to0$ and we find $C_2(\alpha)=\alpha\log\alpha$. Thus $(1)$ is proven and putting $\beta=1$ gives the conclusion $$\int_{0}^{\!\Large \frac{\pi}{2}} x\csc^2(x)\arctan \left(\! \alpha \tan x\right)\, \mathrm{d}x=\frac\pi2\log\left(\frac{(\alpha+1)^{\alpha+1}}{\alpha^\alpha}\right)$$