I am trying to solve following integral.


I started by assuming that $a = \epsilon + \sigma$ and $$\implies x^n = A\tan^2(\theta)$$ Then, $$\implies x = a^{1/n}\tan^{2/n}(\theta)$$ $$\implies dx = \frac{2a^{1/n}}{n} (\tan(\theta))^{2/n-1}\sec^2(\theta)\text{ d}\theta$$

Using $\tan^2(\theta)+1 = \sec^2(\theta)$ and substituing above equations, we can write Eq. 1 as:

$$=\frac{\frac{2a^{2/n}}{n}(\tan(\theta))^{4/n-1}\sec^2(\theta)\text{ d}\theta}{\sec^2(\theta)}$$

$$=\int_0^{\pi/2}\frac{2a^{2/n}}{n}(\tan(\theta))^{4/n-1} \text{ d}\theta$$

let $B = \frac{2a^{2/n}}{n}$, then

$$=\int_0^{\pi/2}B(\tan(\theta))^{4/n-1} \text{ d}\theta$$

Now, I don't know how to proceed forward

The Final answer by authors is: $\large \frac{\pi\sigma(\epsilon+\sigma)^{2/n-1}}{n\sin(\frac{2\pi}{n})}$

  • $\begingroup$ Would you check the final answer please. I think it's $\,\displaystyle\frac{\pi\,(\epsilon+\sigma)^{2/n-1}}{n\,\sin(\frac{2\pi}{n})}\,$ not $\,\displaystyle\frac{\pi\,\color{red}{\sigma}\,(\epsilon+\sigma)^{2/n-1}}{n\,\sin(\frac{2\pi}{n})}\,$. $\endgroup$ – Hazem Orabi Sep 27 '18 at 7:14
  • $\begingroup$ Yes, you are right. Sorry for the typo. $\endgroup$ – Kashan Sep 27 '18 at 21:44

Hint :

Notice that on substitution $x=(at)^{1/n}$ where $a=\epsilon +\sigma$ the integral changes to $$I= \frac {a^{\frac 2n -1}}{n} \int_0^{\infty} \frac {t^{\frac 2n -1}}{1+t}dt $$

Notice that the integral part is just $B\left( \frac 2n, 1-\frac 2n\right)$ Where $B(x,y)$ is standard Beta Function.

Using relation between Gamma function and Beta function and then using the Euler's Reflection formula for the Gamma function i.e. $$\Gamma(z)\Gamma(1-z)=\frac {\pi}{\sin (\pi z)}$$ you could evaluate the integral.

  • $\begingroup$ I was typing a solution using the hypergeometric function when came your answer which is simpler. $\to +1$. Cheers. $\endgroup$ – Claude Leibovici Sep 27 '18 at 7:32
  • $\begingroup$ @Claude Leibovici Thanks :-) $\endgroup$ – Rohan Shinde Sep 27 '18 at 14:20

Observe we have \begin{align} \int^\infty_0 \frac{x}{x^n+a}\ dx = a^{(2-n)/n}\int^\infty_0 \frac{y}{1+y^n}\ dy. \end{align} Hence it suffices to evaluate \begin{align} \int^\infty_0 \frac{y}{1+y^n}\ dy. \end{align} The easiest way is to use contour integration. Observe \begin{align} \int^R_0 \frac{y}{1+y^n}\ dy + \int_{C_R} \frac{z}{1+z^n}\ dz - \int_0^R \frac{(R-t)e^{i4\pi/n}}{1+(R-t)^n}dt = 2\pi i \operatorname{Res}\left(\frac{z}{1+z^n}, z_0=e^{i\pi/n} \right) \end{align} where $C_R$ is the arc of the circle of radius $R$ and angle $2\pi/n$. Hence we see that \begin{align} (1-e^{i4\pi/n}) \int^R_0 \frac{x}{1+x^n}\ dx +\int_{C_R} \frac{z}{1+z^n}\ dz= 2\pi i\frac{z_0}{nz_0^{n-1}} = \frac{2\pi i}{ne^{i\pi (n-2)/n}} \end{align} which means \begin{align} \int^\infty_0 \frac{x}{1+x^n}\ dx =& \frac{2\pi i}{ne^{i\pi(n-2)/n}(1-e^{4i\pi/n})}= \frac{2\pi i}{ne^{i\pi }(e^{-2\pi i/n}-e^{2\pi i/n})} \\ =& \frac{\pi}{n\sin\left(\frac{2\pi}{n} \right)}. \end{align} Finally, combining everything yields \begin{align} \int^\infty_0 \frac{x}{x^n+(\sigma+\varepsilon)}\ dx = \frac{\pi (\sigma+\varepsilon)^{2/n-1}}{n\sin\left(\frac{2\pi}{n} \right)}. \end{align}


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.