2022-01-03

Proof that:
${\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(1+\alpha {\mathrm{sin}}^{2}x\right)dx=\pi \mathrm{ln}\frac{1+\sqrt{1+\alpha }}{2}$

Carl Swisher

Let $I\left(\alpha \right)={\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(1+\alpha {\mathrm{sin}}^{2}x\right)dx$
Then differentiating under the integral sign,
${I}^{\prime }\left(a\right)={\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{sin}}^{2}x}{1+a{\mathrm{sin}}^{2}x}dx={\int }_{0}^{\frac{\pi }{2}}\frac{1}{a+{\mathrm{csc}}^{2}x}dx$
Now let $u=\mathrm{cot}x$
Then
${I}^{\prime }\left(a\right)={\int }_{0}^{\mathrm{\infty }}\frac{1}{a+1+{u}^{2}}\frac{1}{1+{u}^{2}}du$
$=\frac{1}{a}{\int }_{0}^{\mathrm{\infty }}\left(\frac{1}{1+{u}^{2}}-\frac{1}{1+a+{u}^{2}}\right)du$
$=\frac{1}{a}\left(\frac{\pi }{2}-\frac{1}{1+a}{\int }_{0}^{\mathrm{\infty }}\frac{1}{1+\frac{{u}^{2}}{1+a}}du\right)$
$=\frac{1}{a}\left(\frac{\pi }{2}-\frac{1}{\sqrt{1+a}}{\int }_{0}^{\mathrm{\infty }}\frac{1}{1+{v}^{2}}dv\right)$
$=\frac{\pi }{2a}\left(1-\frac{1}{\sqrt{1+a}}\right)$
Then integrating back,
$I\left(\alpha \right)=\frac{\pi }{2}\int \frac{1}{a}\left(1-\frac{1}{\sqrt{1+a}}\right)da$
$=\frac{\pi }{2}\int \frac{1}{{u}^{2}-1}\left(1-\frac{1}{u}\right)2udu$
$=\pi \int \frac{1}{1+u}du$
$=\pi \mathrm{ln}\left(1+\sqrt{1+a}+C$
And since
Therefore,
$I\left(a\right)=\pi \mathrm{ln}\left(\frac{1+\sqrt{1+a}}{2}\right)$

Wendy Boykin

This is quite similar to Random Variable's solution, just the starting integral is different to make the calculations a bit simpler.
Consider
$I\left(b\right)={\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left({b}^{2}+{\mathrm{sin}}^{2}x\right)dx$
$⇒{I}^{\prime }\left(b\right)={\int }_{0}^{\frac{\pi }{2}}\frac{2b}{{b}^{2}+{\mathrm{sin}}^{2}x}dx=2b{\int }_{0}^{\frac{\pi }{2}}\frac{dx}{{b}^{2}+{\mathrm{cos}}^{2}x}$
Factor out ${\mathrm{cos}}^{2}x$ from the denominator and rewrite ${\mathrm{sec}}^{2}x=1+{\mathrm{tan}}^{2}x$ to obtain:
${I}^{\prime }\left(b\right)=2b{\int }_{0}^{\frac{\pi }{2}}\frac{{\mathrm{sec}}^{2}xdx}{{b}^{2}+1+{b}^{2}{\mathrm{tan}}^{2}x}dx$
Use the substitution $\mathrm{tan}x=t$ and evaluating the resulting integral is easy so
${I}^{\prime }\left(b\right)=\frac{\pi }{\sqrt{1+{b}^{2}}}⇒I\left(b\right)=\pi \mathrm{ln}\left(b+\sqrt{1+{b}^{2}}\right)+C$
For , hence $C=-\pi \mathrm{ln}2$
$⇒{\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left({b}^{2}+{\mathrm{sin}}^{2}x\right)dx=\pi \mathrm{ln}\left(\frac{b+\sqrt{1+{b}^{2}}}{2}\right)$
Replace b with $\frac{1}{\sqrt{\alpha }}$ and you get:
${\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(1+\alpha {\mathrm{sin}}^{2}x\right)dx-\frac{\pi }{2}\mathrm{ln}\alpha =\pi \mathrm{ln}\left(\frac{1+\sqrt{1+\alpha }}{2\sqrt{\alpha }}\right)$
$⇒{\int }_{0}^{\frac{\pi }{2}}\mathrm{ln}\left(1+\alpha {\mathrm{sin}}^{2}x\right)dx=\pi \mathrm{ln}\left(\frac{1+\sqrt{1+\alpha }}{2}\right)$

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