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I was trying to solve the integral $\int_0^{\frac{\pi}{2}} \ln(1+\alpha\sin^2 x)\, dx$, where $\alpha>-1$

And my approach is using infinite sum by expanding $\ln(1+\alpha\sin^2 x)$

So let $I=\int_0^{\frac{\pi}{2}} \ln(1+\alpha\sin^2 x)\, dx$ \begin{align*} I&=\int_0^{\pi/2} \sum_{k=0}^{\infty}\frac{(-1)^{k}\alpha^{k+1}\sin^{2k+2} x}{k+1}dx\\ &=\sum_{k=0}^{\infty}\frac{(-1)^{k}\alpha^{k+1}}{k+1} \int_0^{\frac{\pi}{2}} \sin^{2k+2} x dx\\ &=\sum_{k=0}^{\infty}\frac{(-1)^{k}\alpha^{k+1}}{k+1}\frac{(2k+1)!!}{(2k+2)!!}\frac{\pi}{2} \end{align*} I'm stuck at this summation

Edit 1: I'm interested in solving the sum, please do not suggest to solve the integral the other way around as in the link I provided already has sufficient of them

Edit 2: I evaluated $\int_0^{\frac{\pi}{2}} \sin^{2k+2} x dx$ wrong. I also add the domain for $\alpha$.

user635988
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    What makes you think that this approach works? – KBS Sep 16 '22 at 06:15
  • @KBS I'm just curious. Also I believe $\frac{(-\alpha)^{k+1}\sin^{2k+2} x}{k+1}$ is uniformly converge to 0 thus the interchanging of summation and integral is allowable. If we input it into Wolfram, we can actually get an answer. The same as in the solution to the initial problem. But yeah, it seems we have to put a domain to $\alpha$, which I forgot. – user635988 Sep 16 '22 at 06:39

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See https://en.wikipedia.org/wiki/Taylor_series Where $$(1+x)^{-1/2}-1 =\sum_{k=1}^{\infty} {2k \choose k} (\frac{-x}{4})^k=F(x), \quad |x|<1.$$ $$\frac{2I}{\pi}=\sum_{k=0}^{\infty} \frac{(-1)^{k}(2k+1)!!}{(k+1)(2k+2)!!}a^{k+1}=- \sum_{k=0}^{\infty} {2k+2 \choose k+1}\frac{(-a/4)^{k+1}}{k+1}=-\sum_{k=1}^{\infty}{2k \choose k}\frac{(-a/4)^{k}}{k}. $$ $$\frac{2}{\pi}I=-\int_{0}^{a} \frac{F(z)}{z} dz=-\int_{0}^{a}\frac{1-\sqrt{1+z}}{z\sqrt{1+z}}dz=2\int_{\infty}^{1/\sqrt{a}}\left(\frac{1}{\sqrt{1+u^2}}-\frac{1}{u}\right) du, z=1/u^2$$ $$\implies I =\pi\ln \left({1+\sqrt{1+1/u^2}}\right)_{\infty}^{1/\sqrt{a}}=\pi\ln\frac{1+\sqrt{1+a}}{2}.$$

Z Ahmed
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