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I'm struggling with this integral, I've tried different substitutions but i can't solve it.

$$\displaystyle \int_0^{\frac{\pi}{2}}\ln(1+\sin(t)^2)\mathrm{d}t$$

I've tried to calculate both the primitive and the integral using WolframAlpha
But in the primitive appear the polylogarithmic functions with complex argument and for the integral it gives me a numerical approximation ($\approx 0.591331$)

Is there anyone able to solve it? Thanks in advance for your help.

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  • $\begingroup$ I don't think so, how do you turn that formula into my integral? (apart from putting b=1) $\endgroup$
    – user1176548
    May 11 at 9:33
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    $\begingroup$ $a=1$ and $b=\sqrt{2}$ does the job. $\endgroup$
    – TheSilverDoe
    May 11 at 9:35

1 Answer 1

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Using Richard Feynman's integral trick, let: $$f(x):=\int_0^{\frac{\pi}{2}}\ln\left(1+x\cdot \sin(t)^2\right)\mathrm{d}t\\\Rightarrow\;f'(x)=\int_0^{\frac{\pi}{2}}\frac{\sin(t)^2}{1+x\sin(t)^2}\mathrm{d}t=\frac{\pi}{2x}-\frac{1}{x}\int_{0}^{\frac{\pi}{2}}\frac{1}{1+x\sin\left(t\right)^{2}}dt$$ $$\sin(t)^2=s\;\Rightarrow\; t=\arctan(\sqrt{s})\;\Rightarrow\; \mathrm{d}t=\frac{\mathrm{d}s}{2\sqrt{s(1-s)}}$$ $$f'(x)=\frac{\pi}{2x}-\frac{1}{2x}\int_{0}^{1}\frac{1}{1+xs}\cdot\frac{1}{\sqrt{s\left(1-s\right)}}\mathrm{d}s$$ $$f'(x)=\frac{\pi}{2x}-\frac{1}{2x}\frac{\pi}{\sqrt{x+1}}\Rightarrow\; f(x)=\pi\ln\left(\sqrt{x+1}+1\right)+C$$ $$\text{Since }f(0)=0\Rightarrow\; C=-\pi\ln\left(2\right)$$ $$\text{So }f(x)=\pi\ln\left(\frac{1+\sqrt{1+x}}{2}\right)$$ $$\int_0^{\frac{\pi}{2}}\ln\left(1+\sin(t)^2\right)\mathrm{d}t=f(1)=\pi\ln\left(\frac{1+\sqrt{2}}{2}\right)\approx 0.5913306957$$

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