Integral $\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}$

Question

$$ I_1:=\int_0^{\pi/2} \ln(1+\alpha\sin^2 x)\, dx=\pi \ln \frac{1+\sqrt{1+\alpha}}{2}, \qquad \alpha \geq -1. $$ I am trying to prove this integral $I_1$. We can write $$ \int_0^{\pi/2} \ln(\alpha(1/\alpha+\sin^2 x))dx=\int_0^{\pi/2} \left(\ln \alpha+\ln (\frac{1}{\alpha}+\sin^2 x)\right)dx=\frac{\pi}{2} \ln \alpha+I_2 $$ where $$ I_2=\int_0^{\pi/2}\ln (\frac{1}{\alpha}+\sin^2 x) \,dx $$ however I am not sure what that will do for us.... I also tried differentiating wrt $\alpha$ but didn't get placed. How can we prove $I_1$ result? Thanks

Answer 1

Let $ \displaystyle I(a) = \int_{0}^{\pi /2} \ln(1+ a \sin^{2}x) \, dx$.

Then differentiating under the integral sign, $$I'(a) = \int_{0}^{\pi /2} \frac{\sin^{2} x}{1+a \sin^{2} x} \, dx = \int_{0}^{\pi /2} \frac{1}{a+ \csc^{2} x} \, dx .$$

Now let $u = \cot x$.

Then

$$ \begin{align} I'(a) &= \int_{0}^{\infty} \frac{1}{a+1+u^{2}} \frac{1}{1+u^{2}} \, du \\ &= \frac{1}{a} \int_{0}^{\infty} \left(\frac{1}{1+u^{2}} - \frac{1}{1+a+u^{2}} \right) \, du \\ &= \frac{1}{a} \left(\frac{\pi}{2} - \frac{1}{1+a} \int_{0}^{\infty} \frac{1}{1+\frac{u^{2}}{1+a}} \, du \right) \\ &=\frac{1}{a} \left(\frac{\pi}{2} - \frac{1}{\sqrt{1+a}} \int_{0}^{\infty} \frac{1}{1+v^{2}} \, dv \right) \\ &= \frac{\pi}{2a} \left(1 - \frac{1}{\sqrt{1+a}} \right). \end{align}$$

Then integrating back,

$$ \begin{align} I(a) &= \frac{\pi}{2} \int \frac{1}{a} \left(1 - \frac{1}{\sqrt{1+a}} \right) \, da \\ &= \frac{\pi}{2} \int \frac{1}{u^{2}-1} \left(1 - \frac{1}{u} \right) 2u \, du \\ &= \pi \int \frac{1}{1+u} \, du \\ &= \pi \ln \left(1+ \sqrt{1+a} \right) + C. \end{align}$$

And since $I(0) = 0$, $C = -\pi \ln 2$.

Therefore,

$$I(a) = \pi \ln \left(\frac{1 +\sqrt{1+a}}{2} \right) .$$

Answer 2

This is quite similar to Random Variable's solution, just the starting integral is different to make the calculations a bit simpler.

Consider $$I(b)=\int_0^{\pi/2} \ln(b^2+\sin^2x)\,dx$$ $$\Rightarrow I'(b)=\int_0^{\pi/2} \frac{2b}{b^2+\sin^2x}\,dx=2b\int_0^{\pi/2} \frac{dx}{b^2+\cos^2x}$$ Factor out $\cos^2x$ from the denominator and rewrite $\sec^2x=1+\tan^2x$ to obtain: $$I'(b)=2b\int_0^{\pi/2} \frac{\sec^2x\,dx}{b^2+1+b^2\tan^2x}\,dx$$ Use the substitution $\tan x=t$ and evaluating the resulting integral is easy so $$I'(b)=\frac{\pi}{\sqrt{1+b^2}} \Rightarrow I(b)=\pi\ln\left(b+\sqrt{1+b^2}\right)+C$$ For $b=0$, $I(0)=-\pi\ln 2$, hence $C=-\pi\ln2$ $$\Rightarrow \int_0^{\pi/2} \ln(b^2+\sin^2x)\,dx=\pi\ln\left(\frac{b+\sqrt{1+b^2}}{2}\right)$$ Replace $b$ with $1/\sqrt{\alpha}$ and you get: $$\int_0^{\pi/2} \ln(1+\alpha \sin^2x)\,dx-\frac{\pi}{2}\ln \alpha=\pi\ln\left(\frac{1+\sqrt{1+\alpha}}{2\sqrt{\alpha}}\right)$$ $$\Rightarrow \int_0^{\pi/2} \ln(1+\alpha \sin^2x)\,dx=\pi\ln\left(\frac{1+\sqrt{1+\alpha}}{2}\right)$$ $\blacksquare$

Answer 3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x =\pi\,\ln\pars{1 + \root{1 + \alpha} \over 2}:\ {\large ?}}$

\begin{align}& \partiald{}{\alpha}\color{#c00000}{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x} =\partiald{}{\alpha}\int_{0}^{\pi/2}\ln\pars{1 + \alpha\cos^{2}\pars{x}}\,\dd x \\[3mm]&=\int_{0}^{\pi/2}{\cos^{2}\pars{x} \over 1 + \alpha\cos^{2}\pars{x}}\,\dd x ={1 \over \alpha}\int_{0}^{\pi/2} {\bracks{1 + \alpha\cos^{2}\pars{x}} - 1 \over 1 + \alpha\cos^{2}\pars{x}}\,\dd x \\[3mm]&={\pi \over 2\alpha}- {1 \over \alpha}\int_{0}^{\pi/2}{\dd x \over 1 + \alpha\cos^{2}\pars{x}} ={\pi \over 2\alpha}- {1 \over \alpha}\int_{0}^{\pi/2}{\sec^{2}\pars{x}\,\dd x\over \tan^{2}\pars{x} + 1 + \alpha} \\[3mm]&={\pi \over 2\alpha} - {1 \over \alpha\root{1 + \alpha}} \int_{0}^{\infty}{\dd t \over t^{2} + 1} ={\pi \over 2}\pars{{1 \over \alpha} - {1 \over \alpha\root{1 + \alpha}}} \end{align}

\begin{align}& \color{#c00000}{\int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x} ={\pi \over 2}\ \overbrace{\int_{0}^{\alpha}\pars{{1 \over t} - {1 \over t\root{1 + t}}}\,\dd t} ^{\ds{\mbox{Set}\ x \equiv 1 + \root{1 + t}\ \imp\ t = x^{2} - 2x}} \\[3mm]&={\pi \over 2}\int_{2}^{1 + \root{1 + a}}{2\,\dd x \over x} \end{align}

$$\color{#66f}{\large% \int_{0}^{\pi/2}\ln\pars{1 + \alpha\sin^{2}\pars{x}}\,\dd x =\pi\,\ln\pars{1 + \root{1 + \alpha} \over 2}} $$

Answer 4

Differentiate $I_2$ (probably easier than $I_1$) with respect to $a$, then use the Weierstrass substitution to transform it into an integral that you can calculate with residues. I will look into it as well.

Edit: you can also integrate by parts to get rid of the log.

Answer 5

Using Feynman’s Technique Integration, we first differentiate the integral w.r.t. $\alpha$ and obtain $$ \begin{aligned} I^{\prime}(\alpha) &=\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{2} x}{1+\alpha \sin ^{2} x} d x \\ &=\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{1+\alpha \sin ^{2} x-1}{1+\alpha \sin ^{2} x} d x \\ &=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{d x}{1+\alpha \sin ^{2} x} \\ &=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{2} x}{\sec ^{2} x+\alpha \tan ^{2} x} d x \\ &=\frac{\pi}{2 \alpha}-\frac{1}{\alpha} \int_{0}^{\infty} \frac{d t}{1+(1+\alpha) t^{2}}, \text { where } t=\tan x \\ &=\frac{\pi}{2 \alpha}-\frac{1}{\alpha \sqrt{1+\alpha}}\left[\tan ^{-1}(\sqrt{1+\alpha }t)\right]_{0}^{\infty} \\ &=\frac{\pi}{2 \alpha}-\frac{\pi}{2 \alpha \sqrt{1+\alpha}} \end{aligned} $$ Integrating give back the integal $$ \begin{aligned} I(\alpha) &=\frac{\pi}{2} \ln |\alpha|-\frac{\pi}{2} \int \frac{d \alpha}{\alpha \sqrt{1+\alpha}} \\ &=\frac{\pi}{2} \ln |\alpha|-\pi\int \frac{d \sqrt{1+\alpha}}{(\sqrt{1+\alpha})^{2}-1}\\ &=\frac{\pi}{2}\left[\ln |\alpha|+\ln \left(\frac{\sqrt{1+\alpha}+1}{\sqrt{1+\alpha}-1}\right) \right ]+C\\ &=\frac{\pi}{2}\left[\ln |\alpha|+\ln \left(\frac{\sqrt{1+\alpha}+1)^{2}}{\mid \alpha \mid}\right) \right]+C\\ &=\pi \ln (\sqrt{1+\alpha}+1)+C \end{aligned} $$As $I(0)=0$ gives the value $C=-\pi\ln2 $ and hence we can conclude that $$ I(\alpha) =\pi \ln (1+\sqrt{1+\alpha})-\pi\ln 2 =\pi \ln \left(\frac{1+\sqrt{1+\alpha}}{2}\right) .$$