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

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??!!

• WolframAlpha gives the answer $\frac{2\pi}{1+\sqrt{5}}$. – Rory Daulton Apr 5 '15 at 11:29
• @RoryDaulton Maybe this is the simple method! :-) – k1.M Apr 5 '15 at 11:30
• Both examples look incredibly amenable to contour integration techniques. – Frank Apr 5 '15 at 11:32
• The function occurred in the integral is even! So we can find the value easily by contour integration method...but i'm not permitted to use complex integration methods! – k1.M Apr 5 '15 at 11:35
• Wolfram's antiderivative only includes $x$ as $x^{1/5}$ and $x^{2/5}$, so your substitution seems to have made the situation better. From there they appear to take on your idea of complex partial fraction decomposition (they get 4 logs and 6 arctans, each of which only involves $x^{1/5}$ or $x^{2/5}$), but when you do that you really have to be careful about your choices of branch cuts. – Ian Apr 5 '15 at 11:38

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

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.)

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 $$Res\left(\frac{z^6}{1+z^{10}},\zeta^i\right)=\lim_{z\to\zeta^i}\frac{z^6(z-\zeta^i)}{1+z^{10}}$$