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Greetings I am trying to find a closed form for: $$I=\int_0^{\frac{\pi}{2}} x^2 \sqrt{\sin x}\,dx$$ If we rewrite the integral as $$I=\int_0^\infty x^2 \sqrt{\frac{1}{\sqrt{1+\cot^2 x}}}\,dx$$ now with $$\cot x =t $$ $$I=\int_0^{\infty} \operatorname{arccot}^2 (x)(1+x^2)^{-\frac{5}{4}}dx$$ and with https://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms $$I=\frac{1}{4i}\int_0^{\infty}\log^2\left(\frac{z-i}{z+i}\right)(1+x^2)^{-\frac{5}{4}} \, dx$$ Now for the $\log$ I thought to expand into power series but since the radius of converge is abit smaller, this fails. Also integrating by parts or combining the initial integral with $\int_0^{\frac{\pi}{2}} x^2 \sqrt{\cos x}\,dx$ wasn't much of a help, could you help me evaluate this integral ?

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    $\begingroup$ Why do you believe that this has a "nice" closed-from solution? $\endgroup$ – Mark Viola Jul 6 '18 at 20:04
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    $\begingroup$ I just hope for now and well the integral is pretty breezy. $\endgroup$ – Zacky Jul 6 '18 at 20:10
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    $\begingroup$ Wolframalpha doesn't have a closed form, which makes me worried that this will lack a closed form. $\endgroup$ – Chickenmancer Jul 6 '18 at 20:38
  • $\begingroup$ I have no idea what "pretty breezy" means. $\endgroup$ – Mark Viola Jul 6 '18 at 20:50
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    $\begingroup$ @MarkViola It means smooth like butter, clean and crisp. Or not. Actually, probably what's meant is that it looks elementary enough that it should have a closed-form solution. I'd suspect that too, but looks can be deceiving ... $\endgroup$ – John Jul 6 '18 at 21:19
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The substitution $\sin(x) = \sqrt{t}$ leads to the expression $$I = \frac{1}{2} \int \limits_0^1 \frac{t^{-1/4} \arcsin^2 (\sqrt{t})}{\sqrt{1-t}} \, \mathrm{d} t \, . $$ Now you can use the power series for $\arcsin^2$ (see for example this question) and integrate term by term (monotone convergence). Using the beta function you will find \begin{align} I &= \frac{1}{4} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \int \limits_0^1 t^{n-\frac{1}{4}} (1-t)^{-\frac{1}{2}} \, \mathrm{d} t \\ &= \frac{1}{4} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \operatorname{B}\left(n+\frac{3}{4},\frac{1}{2}\right) \\ &= \frac{\sqrt{\pi}}{4} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \frac{\Gamma\left(n+\frac{3}{4}\right)}{\Gamma\left(n+\frac{5}{4}\right)} \\ &= \frac{\sqrt{\pi} \, \Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (2n-1)!!} \frac{\prod_{k=1}^n (4k-1)}{\prod_{l=1}^{n+1} (4l-3)} \\ &= \frac{\pi \sqrt{2 \pi}}{\Gamma\left(\frac{1}{4}\right)^2} \sum \limits_{n=1}^\infty \frac{(2n)!!}{n^2 (4n+1) (2n-1)!!} \prod \limits_{k=1}^n \frac{4k-1}{4k-3} \, . \end{align} Mathematica gives the following expression in terms of a hypergeometric function: $$ I = \frac{6 \pi \sqrt{2 \pi}}{5 \Gamma\left(\frac{1}{4}\right)^2} \, {}_4 \! \operatorname{F}_3 \left(1,1,1,\frac{7}{4};\frac{3}{2},2,\frac{9}{4};1\right) \approx 1.208656578687 \, .$$ Inverse symbolic calculators do not seem to give any expression for this number, so this might be as good as it gets.

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The original integral in the question unfortunately lacks a closed form. You can rewrite it in terms of the hypergeometric function, but that's the best you can get. However I did a bit of digging and found an explicit expression for the integral below

$$\mathfrak{I}=\int\limits_0^{\pi/2}\mathrm dx\, x^2\sqrt{\cos x}$$

Perhaps this may help in some way. First, let $x=\arcsin\sqrt t$ and our integral transforms into

$$\mathfrak{I}=\frac 12\int\limits_0^1\mathrm dt\,t^{-1/2}(1-t)^{-1/4}\arcsin^2\sqrt t$$

The integrand is very similar to the beta function, so we can substitute $\arcsin(\cdot)$ with it's infinite series counterpart and rewrite the integral in terms of the gamma function. Recalling that

$$\arcsin^2z=\sum\limits_{n\geq1}\frac {2^{2n-1}z^{2n}}{n^2\binom {2n}n}$$

We get

$$\begin{align*}\mathfrak{I} & =\frac 14\sum\limits_{n\geq1}\frac {4^n}{n^2\binom {2n}n}\int\limits_0^1\mathrm dt\, t^{n-1/2}(1-t)^{-1/4}\\ & =\frac 14\sum\limits_{n\geq1}\frac {4^n\Gamma^2(n+1)}{n^2\Gamma(2n+1)}\frac {\Gamma\left(n+\frac 12\right)\Gamma\left(\frac 34\right)}{\Gamma\left(n+\frac 54\right)}\end{align*}$$

Using the two equations$$\Gamma\left(n+\frac 12\right)=\frac {\Gamma(2n+1)\sqrt\pi}{4^n\Gamma(n+1)}$$$$\Gamma\left(\frac 14\right)\Gamma\left(\frac 34\right)=\pi\sqrt2$$

The infinite sum simplifies into something much less daunting

$$\mathfrak{I}=\frac {\pi\sqrt{2\pi}}{4\Gamma\left(\frac 14\right)}\sum\limits_{n\geq1}\frac {\Gamma(n)}{n\Gamma\left(n+\frac 54\right)}$$

The gamma functions can again be rewritten as the beta function and converted back to an integral to get

$$\begin{align*}\mathfrak{I} & =\frac {\pi\sqrt{2\pi}}{\Gamma^2\left(\frac 14\right)}\sum\limits_{n\geq1}\frac 1n\operatorname{B}\left(n,\frac 54\right)\\ & =\frac {\pi\sqrt{2\pi}}{\Gamma^2\left(\frac 14\right)}\int\limits_0^1\mathrm dt\,\frac {(1-t)^{1/4}}t\sum\limits_{n\geq1}\frac {t^n}n\\ & =-\frac {\pi\sqrt{2\pi}}{\Gamma^2\left(\frac 14\right)}\int\limits_0^1\mathrm dt\,\frac {(1-t)^{1/4}\log(1-t)}t\end{align*}$$

The resulting integral is rather easy to compute by first using the substitution $1-t\mapsto t^4$ and then using partial fraction decomposition to arrive at

$$\int\limits_0^1\mathrm dt\,\frac {(1-t)^{1/4}\log(1-t)}t=16-8G-\pi^2$$

where $G$ is Catalan's constant. Therefore, it's easy to see that the final result, in color, is

$$\mathfrak{I}=\int\limits_0^{\pi/2}\mathrm dx\, x^2\sqrt{\sin x}\color{blue}{=\frac {\pi\sqrt{2\pi}}{\Gamma^2\left(\frac 14\right)}\biggr[\pi^2+8G-16\biggr]}$$

If we put that together into the original integral, another possible solution, given by ComplexYetTrivial, is $$\int\limits_0^{\pi/2}\mathrm dx\, x^2\sqrt{\sin x}\color{red}{=\frac{\pi\sqrt{2\pi}}{\Gamma^2\left(\frac 14\right)}\biggr[\frac {3\pi^2}2+8G-16\biggr]-\frac {4\pi}{15}{}_3F_2\left[\begin{array}{c}1,\frac 54,\frac 54\\\frac 74,\frac 94\end{array}\right]}$$

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    $\begingroup$ Note that the substitution at the beginning transforms the original integral into $I = \int_0^{\pi/2} (\color{\red}{\frac{\pi}{2}} - x)^2 \sqrt{\cos(x)} \, \mathrm{d} x$, so $I \neq \mathfrak{I}$ . The integral with the constant factor is easy and the one with the quadratic factor is the result of your nice calculation. The part with the linear term does not seem to have a simple closed form, however. According to Mathematica it can be expressed in terms of the hypergeometric function ${}_3 \mathrm{F}_2$ , which is nicer than my own result. I do not know if it can be simplified any further. $\endgroup$ – ComplexYetTrivial Jul 7 '18 at 6:32
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    $\begingroup$ The result should then be $I = \frac{\pi \sqrt{2 \pi}}{\Gamma(\frac{1}{4})^2}[\frac{3}{2} \pi^2 + 8 \mathrm{G} -16] - \frac{4}{15} \pi \, {}_3 \mathrm{F}_2 (1,\frac{5}{4},\frac{5}{4};\frac{7}{4},\frac{9}{4};1)$ . $\endgroup$ – ComplexYetTrivial Jul 7 '18 at 6:41
  • $\begingroup$ @ComplexYetTrivial Oh my god you’re right... I will leave this answer here as a display of my intelligence $\endgroup$ – Frank W. Jul 7 '18 at 13:35

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