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$$\int_{0}^{\infty} \frac{1}{\sqrt{1+x^{10}}} dx$$

I know the answer is $\Gamma(\frac{2}{5})\Gamma(\frac{11}{10})\frac{1}{\sqrt{\pi}}$

Thanks in advance!

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    If you let $u=x^{10}$ you will get a [beta function](https://en.wikipedia.org/wiki/Beta_function). Does that help? – 2017-02-26
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    No, I still can't get it... – 2017-02-26

1 Answers 1

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Beta's function has the form

$$B(a,b)=\int_0^\infty \frac{x^{a-1}}{(1+x)^{a+b}}dx$$

Making the change $u=x^{10}$ (and $du=10x^9dx$) in your integral, you get

$$I=\frac{1}{10}\int_0^\infty \frac{u^{-9/10}}{(1+u)^{1/2}}du=\frac{1}{10}B\left(\frac{1}{10},\frac{2}{5}\right)$$

But

$$B(a,b)=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$

so

$$I=\frac{1}{10}\frac{\Gamma(1/10)\Gamma(2/5)}{\Gamma(1/2)}=\frac{\Gamma(11/10)\Gamma(2/5)}{\sqrt{\pi}}.$$