I'm looking for different methods to solve the following integral.
$$ F\left(\alpha\right) = \int_{0}^{1} x^\alpha \arcsin(x)\:dx$$
For $\alpha > 0$
Here the method I took was to employ integration by parts and then call to special functions, but can this equally be achieved with say a Feynman Trick? or another form integral transform?
My approach in detail:
Employ integration by parts:
\begin{align}
v'(x) &= x^\alpha & u(x) &= \arcsin(x) \\
v(x) &= \frac{x^{\alpha + 1}}{\alpha + 1} & u'(x) &= \frac{1}{\sqrt{1 – x^2}}
\end{align}
Thus,
\begin{align}
F\left(\alpha\right) &= \left[\frac{x^{\alpha + 1}}{\alpha + 1}\cdot\arcsin(x)\right]_0^1 – \int_0^1 \frac{x^{\alpha + 1}}{\alpha + 1} \cdot \frac{1}{\sqrt{1 – x^2}} \:dx \\
&= \frac{\pi}{2\left(\alpha + 1\right)} – \frac{1}{\alpha + 1}\int_0^1 x^{\alpha + 1}\left(1 – x^2\right)^{-\frac{1}{2}} \:dx
\end{align}
Here make the substitution $u = x^2$ to obtain
\begin{align}
F\left(\alpha\right) &= \frac{\pi}{2\left(\alpha + 1\right)} – \frac{1}{\alpha + 1}\int_0^1 \left(\sqrt{u}\right)^{\alpha + 1}\left(1 – u\right)^{-\frac{1}{2}} \frac{\:du}{2\sqrt{u}} \\
&= \frac{\pi}{2\left(\alpha + 1\right)} – \frac{1}{2\left(\alpha + 1\right)}\int_0^1 u^{\frac{\alpha}{2}}\left(1 – u\right) ^{-\frac{1}{2}} \:du \\
&= \frac{1}{2\left(\alpha + 1\right)} \left[ \pi – B\left(\frac{\alpha + 2}{2}, \frac{1}{2} \right) \right]
\end{align}
\begin{align}
F\left(\alpha\right) &=\frac{1}{2\left(\alpha + 1\right)} \left[ \pi – \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{\alpha + 2}{2} + \frac{1}{2}\right)} \right] \\
&= \frac{1}{2\left(\alpha + 1\right)} \left[ \pi – \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)\sqrt{\pi}}{\Gamma\left(\frac{\alpha + 3}{2}\right) } \right] \\
&= \frac{\sqrt{\pi}}{2\left(\alpha + 1\right)} \left[ \sqrt{\pi} – \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)}{\Gamma\left(\frac{\alpha + 3}{2}\right) } \right]
\end{align}
Edits:
Correction of original limit observation (now removed)
Correction of not stating region of convergence for $\alpha$.
Correction of 1/sqrt to sqrt in final line.
Thanks to those commentators for pointing out.
Best Answer
Here is a method that relies on using a double integral.
Noting the integral converges for $\alpha > -2$, recognising $$\arcsin x = \int_0^x \frac{du}{\sqrt{1 - u^2}},$$ the integral can be rewritten as $$I = \int_0^1 \int_0^x \frac{x^\alpha}{\sqrt{1 - u^2}} \, du dx.$$ On changing the order of integration, one has \begin{equation} \int_0^1 \int_u^1 \frac{x^\alpha}{\sqrt{1 - u^2}} \, dx du. \qquad (*) \end{equation} After performing the $x$-integral we are left with $$I = \frac{1}{\alpha + 1} \int_0^1 \frac{1 - u^{\alpha + 1}}{\sqrt{1 - u^2}} \, du, \quad \alpha \neq -1.$$ Enforcing a substitution of $u \mapsto \sqrt{u}$ results in $$I = \frac{1}{2(\alpha + 1)} \int_0^1 \left (\frac{1}{\sqrt{u(1 - u)}} - \frac{u^{\alpha/2}}{\sqrt{ 1 - u}} \right ) \, du = I_1 - I_2.$$
The first of the integrals is trivial $$I_1 = \frac{1}{2(\alpha + 1)} \int_0^1 \frac{du}{\sqrt{\frac{1}{4} - (u - \frac{1}{2})^2}} = \frac{1}{2(\alpha + 1)} \arcsin (2u - 1) \Big{|}^1_0 = \frac{\pi}{2(\alpha + 1)}.$$
For the second of the integrals, it can be evaluated by writing it in terms of the beta function. Here \begin{align} I_2 &= \int_0^1 \frac{u^{\alpha/2}}{\sqrt{1 - u}} \, du\\ &= \int_0^1 u^{(\alpha/2 + 1) - 1} (1 - u)^{1/2 - 1} \, du\\ &= \text{B} \left (\frac{\alpha}{2} + 1, \frac{1}{2} \right )\\ &= \sqrt{\pi} \, \frac{\Gamma \left (\frac{\alpha + 2}{2} \right )}{\Gamma \left (\frac{\alpha + 3}{2} \right )}.\\ \end{align} Thus $$I = \frac{1}{2(\alpha + 1)} \left [\pi - \sqrt{\pi} \, \frac{\Gamma \left (\frac{\alpha + 2}{2} \right )}{\Gamma \left (\frac{\alpha + 3}{2} \right )} \right ], \qquad \alpha \neq -1.$$
For the case when $\alpha = -1$, the double integral at ($*$) becomes $$I = \int_0^1 \int_u^1 \frac{1}{x\sqrt{1 - u^2}} \, dx du.$$ After performing the $x$-integral which yields a natural logarithm, one has $$I = -\int_0^1 \frac{\ln u}{\sqrt{1 - u^2}} \, du.$$ Enforcing a substitution of $u \mapsto \sin u$ leads to $$I = -\int_0^{\pi/2} \ln (\sin u) \, du. \qquad (**)$$ The integral appearing in ($**$) is quite famous and reasonably (?) well known. Its evaluation can be found either here or here. Thus $$I = \frac{\pi}{2} \ln 2, \qquad \alpha = -1.$$