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I attempted to evaluate this integral but seem to be getting nowhere$$I=\int_0^\pi \ln\left(1+\sin^2(t)\right) dt$$

Wolfram returns the value $I\approx 1.18266$ but was not able to provide a closed form for me. I suspect that one could exist but I'm not sure how to proceed. Any help will be appreciated.

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    $\begingroup$ Wolfram is able to get a closed form, but it's nasty. Unless you have a good reason to need a closed form (of which there are few), you'd be better off sticking with the numerics. $\endgroup$ – parsiad Feb 23 '19 at 0:43
  • $\begingroup$ @parsiad I suspect that maybe it could be simplified by plugging in the bounds and applying dilogarithm identities. I was interested in the integral because it was related to a binomial sum I was messing with. $\endgroup$ – aleden Feb 23 '19 at 0:59
  • $\begingroup$ Hint: use $$I(a)=\int_0^\pi \ln\left(1+a\sin^2t\right) dt.$$ $\endgroup$ – xpaul Feb 23 '19 at 0:59
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Here is a different way to set up that a parameter in order to apply Feynman's trick.$$I=2\int_0^\frac{\pi}{2} \ln\left(1+\sin^2t\right) dt\overset{t=\operatorname{arccot} x}=2\int_0^\infty \frac{\ln(2+x^2)-\ln(1+x^2)}{1+x^2}dx$$ Now let us consider $$I(a)=\int_0^\infty \frac{\ln(a+x^2)-\ln(1+x^2)}{1+x^2}dx\Rightarrow I'(a)=\int_0^\infty \frac{1}{(1+x^2)(a+x^2)}dx$$ $$=\frac{1}{a-1}\left(\int_0^\infty \frac{1}{x^2+1}dx-\int_0^\infty \frac{1}{x^2+a}dx\right)=\frac{1}{a-1}\left(\frac{\pi}{2}-\frac{1}{\sqrt a}\cdot \frac{\pi}{2}\right)=\frac{\pi}{2}\frac{1}{\sqrt a(1+\sqrt a)}$$ We are looking to find $I=2I(2)$, but since $I(1)=0$ we have: $$I=2\left(I(2)-I(1)\right)=2\int_1^2 I'(a)da=\pi \int_1^2 \frac{1}{\sqrt a(1+\sqrt a)}da$$ $$\overset {\large a=x^2}=2\pi\int_1^\sqrt 2 \frac{1}{1+x}dx=2\pi \ln(1+x)\bigg|_1^\sqrt 2=2\pi \ln\frac{1+\sqrt 2}{2}$$

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    $\begingroup$ Nice way to use Feynman's, I didn't think to use that sub at the beginning either. Really makes it alot easier, thank you. $\endgroup$ – aleden Feb 23 '19 at 5:20
  • $\begingroup$ I'm happy that I could help! $\endgroup$ – Zacky Feb 23 '19 at 10:32
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Let $$I(a)=\int_0^\pi \ln\left(1+a\sin^2t\right) dt$$ and \begin{eqnarray*} I'(a)&=&\int_0^\pi \frac{\sin^2t}{1+a\sin^2t} dt=\int_0^\pi\frac{1-\cos2t}{2+a(1-\cos2t)}dt\\ &=&\frac{1}{2}\int_0^{2\pi}\frac{1-\cos2t}{2+a(1-\cos2t)}dt=\frac{1}{2}\int_{|z|=1}\frac{1-\frac12(z+\frac1z)}{2+\frac{a}2(z+\frac1z)}\frac{dz}{iz}\\ &=&\frac{1}{2i}\int_{|z|=1}\frac{2z-(z^2+1)}{[4z+a(z^2+1)]z}dz\\ &=&\frac{1}{2i}2\pi i(\text{Res}(f(z),z=0)+\text{Res}(f(z),z=z_1)\\ &=&\pi\bigg(i\frac1a-\frac1{a\sqrt{1+a}}\bigg)=\frac{\pi}{\sqrt{1+a}(\sqrt{1+a}+1)}. \end{eqnarray*} Here $$ z_1=\frac{2+a-2\sqrt{1+a}}{a}. $$ So $$ I(a)=\int_0^1\frac{\pi}{\sqrt{1+a}(\sqrt{1+a}+1)}da=2\pi\ln\frac{1+\sqrt2}{2}.$$

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  • $\begingroup$ Good work!! I think you forgot a differential at the last integral. $\endgroup$ – manooooh Feb 23 '19 at 2:28
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Let $a=3+2\sqrt{2}=(1+\sqrt2)^2$ so that $a^2-6a+1=0$. If we consider $$ \log(a-e^{i2t})=\log|a-e^{i2t}|+i\text{arg}(a-e^{i2t}), $$ then, $$\begin{align*}\log|a-e^{i2t}|=\frac12\log|a-e^{i2t}|^2&=\frac12\log(a^2+1-2a\cos (2t))=\frac12\log((a-1)^2+4a\sin^2 t)\\&=\frac12\log 4a+\frac12\log(1+\sin^2 t).\end{align*}$$ Thus the real part of the integral $$ I=\int_0^\pi \log(a-e^{i2t})\mathrm dt=\frac12\int_0^{2\pi}\log(a-e^{it})\mathrm dt $$ is equal to $\frac{\pi}2\log(4a)+\frac12\int_0^\pi\log(1+\sin^2 t)\mathrm dt$. This gives $$ \int_0^\pi \log(1+\sin^2 t)\mathrm dt =-\pi\log(4a)+2\Re(I). $$ Now, by mean value theorem for analytic (harmonic) functions, we have $$ I=\pi \log(a-z)|_{z=0}=\pi\log a $$ and it follows $$ \int_0^\pi \log(1+\sin^2 t)\mathrm dt = \pi\log\left(\frac{a}4\right)=2\pi\log\left(\frac{1+\sqrt{2}}{2}\right). $$

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Using the series $$ (1-x)^{-1/2}=\sum_{k=0}^\infty\binom{2k}{k}\frac{x^k}{4^k}\tag1 $$ subtracting $1$, dividing by $x$, and integrating, we get $$ \sum_{k=1}^\infty\binom{2k}{k}\frac{x^k}{k4^k}=2\log(2)-2\log\left(1+(1-x)^{1/2}\right)\tag2 $$ Therefore $$ \begin{align} \int_0^\pi\log\left(1+\sin^2(t)\right)\mathrm{d}t &=\int_0^\pi\sum_{k=1}^\infty(-1)^{k-1}\frac{\sin^{2k}(t)}k\,\mathrm{d}t\tag3\\ &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\frac12\int_0^{2\pi}\left(\frac{e^{it}-e^{-it}}{2i}\right)^{2k}\,\mathrm{d}t\tag4\\ &=\sum_{k=1}^\infty\frac{(-1)^{k-1}}k\pi\frac{\binom{2k}{k}}{4^k}\tag5\\[3pt] &=2\pi\log\left(\frac{1+\sqrt2}2\right)\tag6 \end{align} $$ Explanation:
$(3)$: use power series for $\log(1+x)$
$(4)$: $\sin^2(x)=\left(\frac{e^{ix}-e^{-ix}}{2i}\right)^2$ is even
$(5)$: the constant term, $\binom{2k}{k}4^{-k}$, is the only one that does not vanish in the integral
$(6)$: apply $(2)$

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An alternative derivation by integration by parts and very few extras:

$$ I = \int_{0}^{\pi}\log(1+\sin^2\theta)\,d\theta = 2\int_{0}^{\pi/2}1\cdot\log(1+\sin^2\theta)\,d\theta = \pi\log 2-2\int_{0}^{\pi/2}\frac{2\theta\sin\theta\cos\theta}{1+\sin^2\theta}\,d\theta $$ The last integral would be fairly straightforward to compute if $\theta$ was $\sin\theta$. On the other hand we may exploit the Fourier sine series of the identity function over the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$: $$ \theta = \sum_{k\geq 1}\frac{(-1)^{k+1}}{k}\,\sin(2k\theta) \tag{F}$$ then compute $$ \int_{0}^{\pi/2}\frac{\sin(2k\theta)\sin\theta\cos\theta}{1+\sin^2\theta}\,d\theta =\frac{\pi}{2}(\sqrt{2}-1)^{2k}\tag{R}$$ through a recurrence relation. By $(\text{F})$ and $(\text{R})$ it follows that $$ I = \pi\log 2-2\pi\sum_{k\geq 1}\frac{(-1)^{k+1}}{k}(\sqrt{2}-1)^{2k}=\pi\log 2-2\pi\log(4-2\sqrt{2})$$ $$I=\pi\log\left(\frac{3}{4}+\frac{1}{\sqrt{2}}\right)=\color{blue}{2\pi\log\left(\frac{1+\sqrt{2}}{2}\right)}=1.18266139149\ldots $$ and dilogarithms have been carefully avoided.

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