Improper integral of a rational function:$\int_0^\infty \frac{5t^6}{1+t^{10}}dt$

Solution 1:

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 $$ \int_0^\infty \frac{5t^6}{1+t^{10}}dt=\frac12\int _0^1 x^{-7/10}(1-x)^{7/10-1}dx=\frac{\pi}{2\sin(7\pi/10)}=\frac{\sqrt{5}-1}{2}\pi. $$

Solution 2:

Your integral is $$ \int_0^{\infty} \frac{x^{7/5}}{1+x^2} \frac{dx}{x}. $$ Substituting $u=x^2$, $du/u=2dx/x$, the integral becomes $$ \frac{1}{2}\int_0^{\infty} \frac{u^{7/10}}{1+u} \, du $$ Now, $$ \frac{1}{1+u}= \int_0^{\infty} e^{-(1+u)\alpha} \, d\alpha, $$ and interchanging the order of integration gives $$ \frac{1}{2}\int_0^{\infty} e^{-\alpha} \left( \int_0^{\infty} u^{7/10-1} e^{-\alpha u} \, du \right) \, d\alpha $$ Changing variables in the inner integral shows it has value $\alpha^{-7/10}\Gamma(7/10)$. We then do $$ \frac{\Gamma(7/10)}{2} \int_0^{\infty} \alpha^{3/10-1} e^{-\alpha} \, d\alpha = \frac{\Gamma(7/10)}{2}\Gamma(3/10) = \frac{\pi}{2\sin{(3\pi/10)}}, $$ which you do by whatever arcane trigonometry you have to hand.

On the other hand, there is a more direct way of getting the relation $$ \int_0^{\infty} \frac{x^{s-1}}{1+x} \, dx = \frac{\pi}{\sin{\pi s}}, \quad 0<\Re(s)<1. $$ Split the integral in two at $x=1$: $$ \int_0^{1} \frac{x^{s-1}}{1+x} \, dx + \int_1^{\infty} \frac{x^{s-1}}{1+x} \, dx. $$ Now set $u=1/x$ in the second integral, and we find $$ \int_0^{1} \frac{x^{s-1}}{1+x} \, dx + \int_0^{1} \frac{u^{1-s-1}}{1+u} \, du = \int_0^1 \frac{x^{s-1}+x^{-s}}{1+x} \, dx. $$ And now we use a trick I saw in some of G.H. Hardy's work: expand the denominator in a power series and change the order of integration (we can check this is legal easily enough): $$ \int_0^1 \frac{x^{s-1}+x^{-s}}{1+x} \, dx = \sum_{n=0}^{\infty} (-1)^n \int_0^{\infty} (x^{n+s-1}+x^{n-s}) \, dx $$ Doing the integrals gives $$ \int_0^1 \frac{x^{s-1}+x^{-s}}{1+x} \, dx = \sum_{n=0}^{\infty} (-1)^n \left( \frac{1}{n+s} + \frac{1}{n-s+1} \right) $$ Shifting the second terms by one (which you can check is okay by trucating the sums, we end up with $$ \int_0^{\infty} \frac{x^s}{1+x} \, dx = \frac{1}{s} + \sum_{n=1}^{\infty} (-1)^n \left(\frac{1}{s+n}+\frac{1}{s-n} \right), $$ which is a well-known expression for $\pi\csc{\pi s}$. (Which is actually what Hardy says. You can get it using Fourier series, and so avoid both complex analysis and the Gamma function entirely.)

Solution 3:

With some substitutions and ordinary works on the integral we have $$\int_0^\infty \frac{x^{\frac25}}{1+x^2}dx=\frac52\int_{-\infty}^\infty\frac{t^6}{1+t^{10}}dt$$ For every $R\gt0$ we have $$\int_{C}\frac{z^6}{1+z^{10}}dz=\int_{-R}^R\frac{t^6}{1+t^{10}}dt+\int_{C_R}\frac{z^6}{1+z^{10}}dz$$ where $C_R$ is the contour $Re^{it}$ with $t\in [0,\pi]$ and $C=C_R\cup [-R,R]$ with positive direction.

now by using residue theorem on the LHS of the equation above, and letting $R\to\infty$ we obtain $$\sum_{i=1}^5 Res\left(\frac{z^6}{1+z^{10}},\zeta^i\right)=\int_{-\infty}^\infty\frac{t^6}{1+t^{10}}dt$$ because $\int_{C_R}\frac{z^6}{1+z^{10}}dz=o(R)$.
For calculating residues we can use following limits $$Res\left(\frac{z^6}{1+z^{10}},\zeta^i\right)=\lim_{z\to\zeta^i}\frac{z^6(z-\zeta^i)}{1+z^{10}}$$