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 ?
Integral $\int_0^{\frac{\pi}{2}} x^2 \sqrt{\sin x} \, dx$
closed-formintegration
Related Solutions
It is easy to see that the integral is equivalent to
$$ \begin{align*} \int_0^\infty \frac{1}{x\sqrt{2}+\sqrt{2x^2+1}}\frac{\log x}{\sqrt{1+x^2}}dx &= \sqrt{2}\int_0^\infty \frac{\sqrt{x^2+\frac{1}{2}}-x}{\sqrt{1+x^2}}\log x\; dx\tag{1} \end{align*} $$
This integral is a special case of the following generalised equation:
$$\begin{align*}\mathcal{I}(k) :&= \int_0^\infty \frac{\sqrt{x^2+k^2}-x}{\sqrt{1+x^2}}\log x\; dx \\ &= E'(k)-\left(\frac{1+k^2}{2} \right)K'(k)+\left(k^2 K'(k)-E'(k) \right)\frac{\log k}{2}+\log 2-1 \tag{2}\end{align*}$$
where $K'(k)$ and $E'(k)$ are complementary elliptic integrals of the first and second kind respectively.
Putting $k=\frac{1}{\sqrt{2}}$ in equation $(2)$,
$$ \begin{align*} \mathcal{I}\left(\frac{1}{\sqrt{2}}\right)&=E'\left(\frac{1}{\sqrt{2} }\right)-\frac{3}{4}K'\left(\frac{1}{\sqrt{2}} \right)-\left\{\frac{1}{2} K'\left(\frac{1}{\sqrt{2}} \right)-E'\left(\frac{1}{\sqrt{2}} \right)\right\}\frac{\log 2}{4}+\log 2-1 \end{align*} $$ Using the special values,
$$ \begin{align*} E'\left(\frac{1}{\sqrt2} \right) &= \frac{\Gamma\left(\frac{3}{4} \right)^2}{2\sqrt\pi}+\frac{\sqrt{\pi^3}}{4\Gamma\left(\frac{3}{4} \right)^2}\\ K'\left(\frac{1}{\sqrt2} \right) &= \frac{\sqrt{\pi^3}}{2\Gamma\left(\frac{3}{4} \right)^2} \end{align*} $$
we get
$$ \mathcal{I}\left(\frac{1}{\sqrt{2}}\right)=\frac{1+\log\sqrt[4]2}{2\sqrt{\,\pi}}\Gamma\left(\frac34\right)^2-\frac{\sqrt{\,\pi^3}}8\Gamma\left(\frac34\right)^{-2}+(\log 2-1)\, \tag{3} $$
Putting this in equation $(1)$, we get the answer that Cleo posted.
How to prove Equation $(2)$?
We begin with Proposition 7.1 of "The integrals in Gradshteyn and Ryzhik: Part 16" by Boettner and Moll.
$$\int_0^\infty \frac{\log x}{\sqrt{(1+x^2)(m^2+x^2)}}dx = \frac{1}{2}K'(m)\log m$$
Multiplying both sides by $m$ and integrating from $0$ to $k$:
$$ \begin{align*} \int_0^\infty \frac{\sqrt{x^2+k^2}-x}{\sqrt{1+x^2}}\log x\; dx &= \frac{1}{2}\int_0^k m K'(m)\log(m)\; dm \end{align*} $$
The result follows since
$$\begin{align*} \int m K'(m)\log(m)\; dm &= 2E'(m)-\left(1+m^2 \right)K'(m)+\left(m^2 K'(m)-E'(m) \right)\log m\\ &\quad +\text{constant} \tag{4} \end{align*}$$
One can verify equation $(4)$ easily by differentiating both sides with respect to $m$ and using the identities
$$ \begin{align*} \frac{dE'(k)}{dk}&= \frac{k}{k^{'2}}(K'(k)-E'(k))\\ \frac{dK'(k)}{dk}&= \frac{k^2 K'(k)-E^{'}(k)}{kk^{'2}} \end{align*} $$
Define the function $I(s)$ for $s > 0$ by
$$ I(s) = \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - e^{s}\sqrt{x+1} )}{x+1} \, dx. $$
By observing that $I(\infty) = 0$, we have
$$ I(s) = -\int_{s}^{\infty} I'(t) \, dt. $$
Applying Leibniz's integral rule,
$$ I'(s) = \int_{0}^{\infty} \frac{e^{s}\sqrt{x+1}}{1 + (\sqrt{x} - e^{s}\sqrt{x+1} )^{2}} \, \frac{dx}{x+1}. $$
Now with the substitution $x = \tan^{2}\theta$,
\begin{align*} I'(s) &= \int_{0}^{\frac{\pi}{2}} \frac{\sin\theta}{\cosh s - \sin\theta} = \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos\theta}{\cosh s - \cos\theta}. \end{align*}
Then with the substitution $z = e^{i\theta}$, it follows that
\begin{align*} I'(s) &= \frac{i}{2} \int_{\Gamma} \frac{z^{2} + 1}{z^{2} - 2z \cosh s + 1} \frac{dz}{z}, \end{align*}
where the contour $\Gamma$ denotes the counter-clockwise semicircular arc joining from $-i$ to $i$. Note that we have $z^{2} - 2z \cosh s + 1 = 0$ if and only if $\cosh s = \cos \theta$ if and only if $z = e^{i\theta} = e^{\pm s}$. Thus deforming the contour to the straight line joining from $-i$ to $i$ with infinitesimal indent at the origin, we obtain
\begin{align*} I'(s) &= \frac{i}{2} \left\{ \operatorname{PV}\! \int_{-i}^{i} f(z) \, dz + 2\pi i \operatorname{Res} (f, e^{-s}) + \pi i \operatorname{Res} (f, 0) \right\}, \end{align*}
where $f$ denotes the integrand
$$ f(z) = \frac{z^{2} + 1}{z (z^{2} - 2z \cosh s + 1)}. $$
Proceeding the calculation,
\begin{align*} I'(s) &= \frac{i}{2} \left\{ i \operatorname{PV} \! \int_{-1}^{1} f(ix) \, dx + \pi i - 2 \pi i \coth s \right\} \\ &= - \frac{1}{2} \int_{0}^{1} \{ f(ix) + f(-ix) \} \, dx - \frac{\pi}{2} + \pi \coth s \\ &= \cosh s \int_{0}^{1} \frac{\frac{1}{2} (1 - x^{-2}) }{\{ \frac{1}{2}(x + x^{-1}) \}^{2} + \sinh^{2} s} \, dx - \frac{\pi}{2} + \pi \coth s. \end{align*}
Now with the substitution $u = \frac{1}{2} \{ x + x^{-1} \}$, it follows that
\begin{align*} I'(s) &= - \cosh s \int_{1}^{\infty} \frac{du}{u^{2} + \sinh^{2} s} - \frac{\pi}{2} + \pi \coth s \\ &= - \arctan(\sinh s) \coth s - \frac{\pi}{2} + \pi \coth s. \end{align*}
Finally, with the substitution $x = \sinh t$,
\begin{align*} I(s) = - \int_{s}^{\infty} I'(t) \, dt &= \int_{s}^{\infty} \left\{ \arctan(\sinh t) \coth t + \frac{\pi}{2} - \pi \coth t \right\} \, dt \\ &= \int_{\sinh s}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx. \end{align*}
Our problem corresponds to $s = \log 2$, or equivalently, $a := \sinh s = \frac{3}{4}$.
\begin{align*} &\int_{a}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx \\ &= - \int_{a}^{\infty} \frac{\arctan (1/x)}{x} \, dx + \frac{\pi}{2} \int_{a}^{\infty} \left( \frac{1}{\sqrt{x^{2} + 1}} - \frac{1}{x} \right) \, dx \\ &= - \int_{0}^{1/a} \frac{\arctan x}{x} \, dx + \frac{\pi}{2} \lim_{R\to\infty} \left[ \log\left( \frac{x + \sqrt{x^{2} + 1}}{x} \right) \right]_{a}^{R} \\ &= - \operatorname{Ti}\left(\frac{1}{a}\right) - \frac{\pi}{2} \log\left( \frac{a + \sqrt{a^{2} + 1}}{2a} \right), \end{align*}
where the function
$$ \operatorname{Ti}(x) = \frac{\operatorname{Li}_{2}(ix) - \operatorname{Li}_{2}(-ix)}{2i} $$
is the inverse tangent integral. Using the following simple identity
$$ \operatorname{Ti}(x) = \operatorname{Ti}\left(\frac{1}{x}\right) + \frac{\pi}{2}\log x, $$
it follows that
\begin{align*} \int_{a}^{\infty} \left( \frac{\arctan x}{x} + \frac{\pi}{2\sqrt{x^{2} + 1}} - \frac{\pi}{x} \right) \, dx = - \operatorname{Ti}(a) - \frac{\pi}{2} \log\left( \frac{a + \sqrt{a^{2} + 1}}{2a^{2}} \right). \end{align*}
Therefore, plugging $a = \frac{3}{4}$ gives
$$ \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - 2\sqrt{x+1} )}{x+1} \, dx = \pi \log(3/4) - \operatorname{Ti}(3/4) $$
as Cleo pointed out without explanation. More generally, for $k > 1$
$$ \int_{0}^{\infty} \frac{\operatorname{arccot}( \sqrt{x} - k \sqrt{x+1} )}{x+1} \, dx = \frac{\pi}{2} \log \left( \frac{(k^{2} - 1)^{2}}{2k^{3}} \right) - \operatorname{Ti}\left( \frac{k^{2} - 1}{2k} \right). $$
Best Answer
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.