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Let $x \in \mathbb{R}^{+}$, I wonder how to prove that $$ F\left(x\right)=\int_{0}^{\, \pi/2}\ln\left(1+x\sin^2\left(t\right)\right)\text{d}t=\pi \ln\left(\frac{1}{2}\left(1+\sqrt{1+x}\right)\right) $$ It would beautifully show that $$ \int_{0}^{\, \pi/2}\ln\left(1+4\sin^2\left(t\right)\right)\text{d}t=\pi \ln\left(\varphi\right) $$ I've no idea how to proceed, except maybe that $$ F\left(x\right)=\pi\ln\left(\sqrt{1+x}\right)-x\int_{0}^{\, \pi/2}\frac{\sin\left(2t\right)}{1+x\sin^2\left(t\right)}\text{d}t $$ Then I guess the second will be the missing part, but how to compute it easily ?

Martin Sleziak
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Atmos
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  • This has been asked before, many times. Efficient ways are Feynman's trick or just trigonometric identities and Riemann sums. Have a look at pages $45+$ of my notes – Jack D'Aurizio Jan 03 '18 at 23:43
  • Related: https://math.stackexchange.com/questions/873846/int-0-pi-2-ln-left14-sin4-x-right-mathrmdx-and-the-golden-ratio – Olivier Oloa Jan 04 '18 at 08:24

2 Answers2

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Following Jack D'Aurizio's idea: differentiate under the integral sign to get: \begin{align} F'(x)=\int^{\pi/2}_0 \frac{\sin^2(t)}{1+x\sin^2(t)}\,dt \end{align} This can be easily done by setting $t=\arctan(z)$: \begin{align} F'(x)=\int^\infty_0 \frac{z^2}{(1+(1+x)z^2)(1+z^2)}\,dz \end{align} Which can be easily solved by partial fraction decomposition or Residue Theorem to get: \begin{align} F'(x)=\frac{\pi}{2+2\sqrt[]{1+x}+2x} \end{align} Integrating this back for example with substitution $1+x=u^2$ you finally get: \begin{align} F(x)=\pi\ln\left(1+\sqrt[]{x+1}\right)+C \end{align} By setting $x=0$ and noticing that $\ln(1)=1$ we get: \begin{align} \pi\ln(2)+C=0 \end{align} Hence $C=\pi\ln(1/2)$. So:

\begin{align} F(x)=\int^{\pi/2}_0 \ln\left(1+x\sin^2(t)\right)\,dt=\pi\ln\left( \frac{1+\sqrt[]{1+x}}{2}\right) \end{align}

Shashi
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Consider

$$ F'(x) =\int_0^{\pi/2} \frac{\sin^2 t}{1+x\sin^2t}dt = \int_0^{\pi/2} \frac{1}{\csc^2t + x}dt $$

Let $u = \cot t$ $$ \begin{align} F'(x) &= \int_0^\infty \frac{1}{(u^2+1)(u^2+1+x)}du \\ &= \frac{1}{x}\int_0^\infty \left(\frac{1}{u^2+1} - \frac{1}{u^2+1+x} \right) du \\ &= \frac{1}{x}\left(\frac{\pi}{2} - \frac{\pi}{2\sqrt{1+x}}\right) \\ &= \frac{\pi}{2\sqrt{1+x}(\sqrt{1+x}+1)} \end{align} $$

Since $F(0)=0$, we have $$ \begin{align} F(x) &= \frac{\pi}{2}\int_0^x \frac{1}{\sqrt{1+t}(\sqrt{1+t}+1)}dt \\ &= \pi \int_1^{\sqrt{1+x}} \frac{1}{1+u}du \\ &= \pi\ln \left(\frac{1 + \sqrt{1+x}}{2}\right) \end{align} $$

Dylan
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