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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.

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    $\begingroup$ How would you expect to get $\Gamma$'s out of Feynman Trick or whatever else? (And, BTW, $\displaystyle\lim_{\alpha\to\infty}F(\alpha)=0$). $\endgroup$ – metamorphy Dec 10 '18 at 1:03
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    $\begingroup$ There needs to be some restrictions on the value of $\alpha$ otherwise the improper integral will not converge, namely $\alpha > -2$. The case $\alpha = -1$ also needs to be handled with care. $\endgroup$ – omegadot Dec 10 '18 at 2:12
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    $\begingroup$ Yes, sorry. I meant to say it only converges for $\alpha > -2$ and diverges for $\alpha \leqslant -2$. $\endgroup$ – omegadot Dec 10 '18 at 3:02
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    $\begingroup$ I'm fairly certain that's the fastest way to do it $\endgroup$ – clathratus Dec 10 '18 at 3:29
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    $\begingroup$ The $\frac{1}{\sqrt{\pi }}$ in the last line needs to be $\sqrt{\pi}$. $\endgroup$ – JimB Dec 10 '18 at 5:12
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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 $$I = \frac{\pi}{2} \ln 2, \qquad \alpha = -1.$$

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  • $\begingroup$ A fantastic solution. Thank you very much very. I'm been working on developing my general nasty integral skills and this mode of converting a single to double integral is great. Is this a form of Feynman's Trick? Regardless, again, fantastic solution. $\endgroup$ – user150203 Dec 10 '18 at 12:10
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Answer 2.0:

We know that for $|x|<1$, $$\arcsin x=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)}x^{2k+1}$$ Where $$(1/2)_k=\frac{\Gamma(1/2+k)}{\Gamma(1/2)}$$ Hence we may begin with $$F(a)=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)}\int_0^1x^{2k+a+1}\mathrm dx=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)(2k+2+a)}$$ Then we play with the fractions a little to get $$F(a)=\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!}\bigg[\frac1{2k+1}-\frac1{2k+2+a}\bigg]$$ $$F(a)=\frac1{a+1}\sum_{n\geq0}\frac{(1/2)_n}{n!}\frac1{2n+1}-\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!}\frac1{2k+2+a}$$ $$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+2+a)}$$ $$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!}\int_0^1x^{2k+1+a}\mathrm dx$$ $$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\int_0^1\sum_{k\geq0}\frac{(1/2)_k}{k!}x^{2k+1+a}\mathrm dx$$ $$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\int_0^1\frac{x^{a+1}}{\Gamma(1/2)}\sum_{k\geq0}\frac{\Gamma(1/2+k)}{k!}x^{2k}\mathrm dx$$ $$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\int_0^1\frac{x^{a+1}}{\sqrt{1-x^2}}\mathrm dx$$ $u=x^2$: $$F(a)=\frac\pi{2(a+1)}-\frac1{2(a+1)}\int_0^1u^{a/2}(1-u)^{-1/2}\mathrm du$$ $$F(a)=\frac\pi{2(a+1)}-\frac{\Gamma(\frac{a+2}2)\Gamma(\frac12)}{2(a+1)\Gamma(\frac{a+3}2)}$$ $$F(a)=\frac{\sqrt{\pi}}{2(a+1)}\bigg[\sqrt{\pi}-\frac{\Gamma(\frac{a+2}2)}{\Gamma(\frac{a+3}2)}\bigg]$$

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