Find the value of the integral
$$\int_0^\infty \frac{x^{\frac25}}{1+x^2}dx.$$
I tried the substitution $x=t^5$ to obtain
$$\int_0^\infty \frac{5t^6}{1+t^{10}}dt.$$
Now we can factor the denominator to polynomials of degree two (because we can easily find all roots of polynomial occured in the denominator of the former integral by using complex numbers) and then by using partial fraction decomposition method find the integral!
Is there any simple method to find the integral value??!!
Best Answer
One may recall the Euler beta function $$ B(a,b) =\int _0^1 x^{a-1}(1-x)^{b-1}dx=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}, $$ with a remarkable case $$ \int _0^1 x^{a-1}(1-x)^{-a}dx=\Gamma(a)\Gamma(1-a)=\frac{\pi}{\sin(\pi a)}. $$
Then, by the change of variable $\displaystyle x=\frac{1}{1+t^{10}}$, giving $\displaystyle t=x^{-1/10}(1-x)^{1/10}$, we get $$ \int_0^\infty \frac{5t^6}{1+t^{10}}dt=\frac12\int _0^1 x^{-7/10}(1-x)^{7/10-1}dx=\frac{\pi}{2\sin(7\pi/10)}=\frac{\sqrt{5}-1}{2}\pi. $$