4
$\begingroup$

I am trying to solve following integral.

$$=\int_0^\infty\frac{x}{x^n+(\epsilon+\sigma)}dx$$

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: $\large \frac{\pi\sigma(\epsilon+\sigma)^{2/n-1}}{n\sin(\frac{2\pi}{n})}$

$\endgroup$
2
  • $\begingroup$ 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 $\,\displaystyle\frac{\pi\,\color{red}{\sigma}\,(\epsilon+\sigma)^{2/n-1}}{n\,\sin(\frac{2\pi}{n})}\,$. $\endgroup$ Commented Sep 27, 2018 at 7:14
  • $\begingroup$ Yes, you are right. Sorry for the typo. $\endgroup$
    – SJa
    Commented Sep 27, 2018 at 21:44

2 Answers 2

5
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ I was typing a solution using the hypergeometric function when came your answer which is simpler. $\to +1$. Cheers. $\endgroup$ Commented Sep 27, 2018 at 7:32
  • $\begingroup$ @Claude Leibovici Thanks :-) $\endgroup$ Commented Sep 27, 2018 at 14:20
3
$\begingroup$

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}

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .