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I'm trying to express this integral $$ \int_0^1\sqrt[2\,n\,]{\frac x{1-x}}\,\mathrm dx $$

for any $n\in\mathbb N$. I've tried integral by substitution and partial fraction decomposition but it does not seem to lead to the solution.

Do you have any advice ?

My point is to calculate this integral: enter image description here

For this I divided it in two integrals:

enter image description here

I used the substitution method twice. Is the result right now ? Was there an easier way to express this integral ?

Thanks a lot !

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This is what is known as a beta function. Generally:

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

In your case,

$$a=1+\frac1{2 n}$$ $$b = 1-\frac1{2 n} $$

So the result is

$$\Gamma \left ( 1+\frac1{2 n} \right ) \Gamma \left ( 1-\frac1{2 n} \right ) $$

which may be simplified to

$$-\frac1{2 n} \Gamma \left ( 1+\frac1{2 n} \right ) \Gamma \left ( -\frac1{2 n} \right ) = \frac{\pi}{2 n \sin{\frac{\pi}{2 n}}}$$

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An alternative method arises by substituting $y=x/(1-x)$; then the integral becomes

$$\int_0^{\infty} dy \, \frac{y^{1/(2 n)} }{(1+y)^2} $$

This may be evaluated by contour integration using the residue theorem. Consider the contour integral

$$\oint_C dz \, \frac{z^{1/(2 n)} }{(1+z)^2} $$

where $C$ is a keyhole contour about the positive real axis. The contour integral may be shown to be equal to

$$\left (1-e^{i \pi/n}\right ) \int_0^{\infty} dx \, \frac{x^{1/(2 n)} }{(1+x)^2} $$

By the residue theorem, the contour integral is also equal to

$$-i 2 \pi \frac1{2 n} e^{i \pi/(2 n)} $$

The integral is thus equal to

$$\int_0^{\infty} dx \, \frac{x^{1/(2 n)} }{(1+x)^2} = \frac{\pi}{2 n \sin{\frac{\pi}{2 n}}} $$

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  • $\begingroup$ Thanks a lot ! I didn't know neither about the beta function nor about the residue theorem, I'll try to remerber it for next time. $\endgroup$ – user225100 Mar 20 '15 at 12:05
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My point was to calculate this integral: enter image description here

For this I divided it in two integrals:

enter image description here

I used the substitution method twice. Is the result right now ? Was there an easier way to express this integral ?

[I don't know if it's the right place to ask an other question... I'm sorry if it's not]

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  • $\begingroup$ Yes, ask another question. Better yet, search the site. $\endgroup$ – Ron Gordon Mar 20 '15 at 12:24

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