# Closed form for $\int_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$

I've been looking at

$$\int\limits_0^\infty {\frac{{{x^n}}}{{1 + {x^m}}}dx }$$

It seems that it always evaluates in terms of $\sin X$ and $\pi$, where $X$ is to be determined. For example:

$$\int\limits_0^\infty {\frac{{{x^1}}}{{1 + {x^3}}}dx = } \frac{\pi }{3}\frac{1}{{\sin \frac{\pi }{3}}} = \frac{{2\pi }}{{3\sqrt 3 }}$$

$$\int\limits_0^\infty {\frac{{{x^1}}}{{1 + {x^4}}}dx = } \frac{\pi }{4}$$

$$\int\limits_0^\infty {\frac{{{x^2}}}{{1 + {x^5}}}dx = } \frac{\pi }{5}\frac{1}{{\sin \frac{{2\pi }}{5}}}$$

So I guess there must be a closed form - the use of $\Gamma(x)\Gamma(1-x)$ first comess to my mind because of the $\dfrac{{\pi x}}{{\sin \pi x}}$ appearing. Note that the arguments are always the ratio of the exponents, like $1/4$, $1/3$ and $2/5$. Is there any way of finding it? I'll work on it and update with any ideas.

The integral reduces to finding

$$\int\limits_{ - \infty }^\infty {\frac{{{e^{a t}}}}{{{e^t} + 1}}dt}$$

With $a =\dfrac{n+1}{m}$ which converges only if

$$0 < a < 1$$

Using series I find the solution is

$$\sum\limits_{k = - \infty }^\infty {\frac{{{{\left( { - 1} \right)}^k}}}{{a + k}}}$$

-
Digamma function argument will work, though unnatural. – Sangchul Lee Feb 17 '12 at 20:57
possible duplicate of Interesting integral formula – Listing Feb 17 '12 at 21:46
@Listing Please, do not close this thread. I know nothing about complex or countour integration and thus I can't use the solutions given in the thread. I'm sticking to series and Digammas. – Pedro Tamaroff Feb 17 '12 at 22:19
@sos440 What do you mean by "unnatural"? I usually use that adjective but I guess it is rather a personal use. See here where I use the Digamma to solve this. – Pedro Tamaroff Feb 21 '12 at 3:36
The solution to this problem is also given here: math.stackexchange.com/questions/48740/int-0-infty-fracdx1xn/… – Eric Naslund Feb 29 '12 at 17:09

I would like to make a supplementary calculation on BR's answer.

Let us first assume that $0 < \mu < \nu$ so that the integral $$\int_{0}^{\infty} \frac{x^{\mu-1}}{1+x^{\nu}} \; dx$$ converges absolutely. By the substitution $x = \tan^{2/\nu} \theta$, we have $$\frac{dx}{1+x^{\nu}} = \frac{2}{\nu} \tan^{(2/\nu)-1} \theta \; d\theta.$$ Thus \begin{align*} \int_{0}^{\infty} \frac{x^{\mu-1}}{1+x^{\nu}} \; dx & = \int_{0}^{\frac{\pi}{2}} \frac{2}{\nu} \tan^{\frac{2\mu}{\nu}-1} \theta \; d\theta \\ & = \frac{1}{\nu} \beta \left( \frac{\mu}{\nu}, 1 - \frac{\mu}{\nu} \right) \\ & = \frac{1}{\nu} \Gamma \left( \frac{\mu}{\nu} \right) \Gamma \left( 1 - \frac{\mu}{\nu} \right) \\ & = \frac{\pi}{\nu} \csc \left( \frac{\pi \mu}{\nu} \right), \end{align*} where the last equality follows from Euler reflexion formula.

-
That's better. Thanks. – Pedro Tamaroff Feb 17 '12 at 20:48
Any info on the series representation? – Pedro Tamaroff Feb 17 '12 at 20:55
Check my other question, I have found a way to prove it with the Digamma function – Pedro Tamaroff Feb 18 '12 at 23:47
@sos440: nice way.(+1) – OFFSHARING Aug 11 '12 at 12:07

The general formula (for $m > n+1$ and $n \ge 0$) is $\frac{\pi}{m} \csc\left(\frac{\pi (n+1)}{m}\right)$. IIRC the usual method involves a wedge-shaped contour of angle $2 \pi/m$.

EDIT: Consider $\oint_\Gamma f(z)\ dz$ where $f(z) = \frac{z^n}{1+z^m}$ (using the principal branch if $m$ or $n$ is a non-integer) and $\Gamma$ is the closed contour below:

$\Gamma_1$ goes to the right along the real axis from $\epsilon$ to $R$, so $\int_{\Gamma_1} f(z)\ dz = \int_\epsilon^R \frac{x^n\ dx}{1+x^m}$. $\Gamma_3$ comes in along the ray at angle $2 \pi/m$. Since $e^{(2 \pi i/m) m} = 1$, $\int_{\Gamma_3} f(z)\ dz = - e^{2 \pi i (n+1)/m} \int_{\Gamma_1} f(z)\ dz$. $\Gamma_2$ is a circular arc at distance $R$ from the origin. Since $m > n+1$, the integral over it goes to $0$ as $R \to \infty$. Similarly, the integral over the small circular arc at distance $\epsilon$ goes to $0$ as $\epsilon \to 0$. So we get

$$\lim_{R \to \infty, \epsilon \to 0} \int_\Gamma f(z)\ dz = (1 - e^{2 \pi i (n+1)/m}) \int_0^\infty \frac{x^n\ dx}{1+x^m}$$

The meromorphic function $f(z)$ has one singularity inside $\Gamma$, a pole at $z = e^{\pi i/m}$ where the residue is $- e^{\pi i (n+1)/m}/m$. So the residue theorem gives you

$$\int_0^\infty \frac{x^n\ dx}{1+x^m} = \frac{- 2 \pi i e^{\pi i (n+1)/m}}{ m (1 - e^{2 \pi i (n+1)/m})} = \frac{\pi}{m} \csc\left(\frac{\pi(n+1)}{m}\right)$$

-
"$\Gamma_3$ comes in along the ray at angle 2π/m. Since $e^{(2πi/m)m}=1, \int_{\Gamma_3}(z) \,dz=−e^{2πi(n+1)/m}∫_{\Gamma_1}f(z)\, dz$". How is it that the integral over $\Gamma_3$ becomes a multiple of the integral over $\Gamma_1$? – john.abraham Nov 21 '13 at 18:30
Hi @RobertIsrael, I like the method of using a wedge contour - it obviously minimizes computation of residues, and the conversion from complex number (lots of Euler formulas) to real number as a final answer to the original, real integral is a much easier process. However, say there are 3 roots, and a wedge going from 0 argument to pi/2 is enough to enclose 1 of the poles, why do we insist on using a wedge of angle 2pi/3 = 120 degrees? If we use a wedge of angle pi/2, does the positive imaginary axis pose a problem? Thanks Professor Israel, – Lebron James Oct 23 at 5:52
The wedge contour works if the integrals over the radial parts are related by symmetry: multiplying $z$ by $e^{i\theta}$, where $\theta$ is the angle of the wedge, multiplies the function by a constant. The choice of angle is dictated by the symmetry of the function. – Robert Israel Oct 23 at 6:13

Assuming only that $m>0$ and $-1<n<m-1$, let $z=x^m$ with the exaggerated contour $\gamma$:

$\hspace{4.5cm}$

$\gamma$ is actually tight above and below the positive real axis and around the origin and circles back at an arbitrarily large distance from the origin.

The part just above the positive real axis captures the integral. The part just below the positive real axis gets $-e^{2\pi i\frac{n-m+1}{m}}$ times the integral. The residue at $z=-1$ of $\frac{z^{\frac{n-m+1}{m}}}{1+z}$ is $e^{\pi i\frac{n-m+1}{m}}$. Putting this all together yields

$$\frac1m\int_\gamma\frac{z^{\frac{n-m+1}{m}}}{1+z}\,\mathrm{d}z=\left(1-e^{2\pi i\frac{n-m+1}{m}}\right)\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x$$ $$\frac{2\pi i}{m}e^{\pi i\frac{n-m+1}{m}}=\left(1-e^{2\pi i\frac{n-m+1}{m}}\right)\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x$$ $$\frac{\pi}{m}=\sin\left(\pi\frac{n+1}{m}\right)\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x$$ $$\frac{\pi}{m}\csc\left(\pi\frac{n+1}{m}\right)=\int_0^\infty\frac{x^n}{1+x^m}\,\mathrm{d}x$$ Relation to $\Gamma(\alpha)\Gamma(1-\alpha)$

Setting $m=1$, $n=\alpha-1$, and using the substitution $s=tu$ yields Euler's Reflection Formula: \begin{align} \Gamma(\alpha)\Gamma(1-\alpha) &=\int_0^\infty s^{\alpha-1}e^{-s}\,\mathrm{d}s\int_0^\infty t^{-\alpha}e^{-t}\,\mathrm{d}t\\ &=\int_0^\infty\int_0^\infty u^{\alpha-1}e^{-(tu+t)}\,\mathrm{d}u\,\mathrm{d}t\\ &=\int_0^\infty\frac{u^{\alpha-1}}{1+u}\,\mathrm{d}u\\[6pt] &=\pi\csc(\pi\alpha) \end{align}

-
(+1) neat as always! – user153330 Feb 1 at 19:04
Hi @robjohn, when faced with p.v. integrals, I always try to use this keyhole contour as my first method -- but if I remember correctly, this method has only ever worked nicely and correctly...once. Every other time, I've had to use a wedge contour, similar to what Robert Israel shows above. I wanted to ask you: are there times when the keyhole contour is the clearly better choice of contour - perhaps even necessary? If not, then using the wedge contour minimizes computation of residues and seems like the safest method to go with, especially on an exam, with the clock ticking. – Lebron James Oct 23 at 5:43
What do you think? Thanks @robjohn, – Lebron James Oct 23 at 5:43
Hi @robjohn, just a heads up, I've posted this comment as a new question on MSE. Thanks, – Lebron James Oct 23 at 6:13

This is in Gradshteyn/Ryzhik, formula 3.241.2: $$\int_0^\infty {x^{\mu-1}\over 1+x^\nu}\ dx={\pi\over\nu}\csc\bigg({\mu\pi\over\nu}\bigg)={1\over\nu}B\bigg({\mu\over\nu},{\nu-\mu\over\nu}\bigg)$$ assuming, of course, $Re(\nu)>Re(\mu)>0$, and where $B(x,y)$ denotes the Beta function $B(x,y)={\Gamma(x)\Gamma(y)\over\Gamma(x+y)}$.

To see the Beta function part of the equality, use the integral representation $$B(x,y)=\int_0^\infty {t^{x-1}\over (1+t)^{x+y}}\ dt$$ Then ${1\over\nu}B\big({\mu\over\nu},{\nu-\mu\over\nu}\big)$ is $${1\over\nu}\int_0^\infty {t^{{\mu\over\nu}-1}\over 1+t}\ dt$$ Send $t$ to $t^\nu$, and you are done!

-
The Beta function has an integral representation $$B(x,y)=\int_0^\infty {t^{x-1}\over (1+t)^{x+y}}\ dx$$ Since we have the answer, using this formula, it is not hard to show it is correct :) – B R Feb 17 '12 at 20:16
But my integral has $1+x^v$, not $(1+x)^v$. How did you transform it? – Pedro Tamaroff Feb 17 '12 at 20:25
@B R: nice way (+1) – OFFSHARING Aug 11 '12 at 12:06

Use Ramanujan's Master theorem, with $u=x^b\Rightarrow dx=\frac{du}{bu^{1-\frac{1}{b}}}$:

$$\int_0^{\infty}\frac{x^{a-1}}{1+x^b}dx=\frac{1}{b}\int_0^{\infty}\frac{u^{\frac{a-1}{b}-1+\frac{1}{b}}}{1+u}du$$ $$=\frac{1}{b}\int_0^{\infty}u^{\frac{a}{b}-1}\sum_{k \ge 0}\frac{(-u)^k}{\Gamma(k+1)}\Gamma(k+1)du$$ $$=\frac{1}{b}\Gamma\left(\frac{a}{b}\right)\Gamma\left(1-\frac{a}{b}\right)=\frac{\pi}{b \sin \left(\frac{a\pi}{b}\right)}.$$

-

Simpler way to compute a definite integral without resorting to partial fractions?

$\int_{0}^{\infty}\frac{dx}{1+x^n}$

provide a solution to exactly this problem.

Is there an elementary method for evaluating $\int_0^\infty \frac{dx}{x^s (x+1)}$?

The meta thread discussing the similarities between these 4 questions can be found here: Abstract Duplicates? What to do in this scenario.

-
Note that the previous qustions don't ask for the general solution but rather a particular case, and it is the answerers that decide to provide a solution to the general integral I present here. Also, I wanted another approach in the solution, see here. – Pedro Tamaroff Feb 29 '12 at 23:48

$$t=\frac1{1+x^m}\qquad\iff\qquad x=\left(\frac1t-1\right)^\frac1m=\left(\frac{1-t}t\right)^\frac1m=t^{^{-\frac1m}}\cdot(1-t)^{^\frac1m}\qquad\iff$$

Can you already see where this is going ? ;-)

$$\iff\frac{dx}{dt}=\frac{d}{dt}\left[\left(\frac1t-1\right)^\frac1m\ \right]=\frac1m\left(\frac{1-t}t\right)^{\frac1m-1}\left(-\frac1{t^2}\right)=-\frac{t^{^{-\frac1m-1}}\cdot(1-t)^{^{\frac1m-1}}}m\iff$$

$$I=\int_0^\infty\frac{x^n}{1+x^m}dx=\frac1m\int_0^1t^{^{-\frac{n+1}m}}\cdot(1-t)^{^{\frac{n+1}m-1}}dt=\frac{B\left(1-\frac{n+1}m,\frac{n+1}m\right)}m$$

Now all we have to do is use the fact that $B(m,n)=\displaystyle\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$ , along with Euler's reflection formula, $\displaystyle\Gamma(z)\cdot\Gamma(1-z) = \frac{\pi}{\sin{(\pi z)}}$ , and we're done ! :-)

-


\begin{align} &\sum_{k = - \infty }^\infty{\pars{-1}^{k} \over a + k} = {1 \over a} + \sum_{k = 1}^{\infty}\bracks{ {\pars{-1}^{-k} \over a - k} + {\pars{-1}^{k} \over a + k}} = {1 \over a} + \sum_{k = 1}^{\infty}\pars{-1}^{k + 1} {2a \over \pars{k - a}\pars{k + a}} \\[3mm]&= {1 \over a} + \sum_{k = 0}^{\infty}{2a \over \pars{2k + 1 - a}\pars{2k + 1 + a}} - \sum_{k = 0}^{\infty}{2a \over \pars{2k + 2 - a}\pars{2k + 2 + a}} \\[3mm]&= {1 \over a} + {a \over 2} \sum_{k = 0}^{\infty}{1 \over \pars{k + \pars{1 - a}/2}\pars{k + \pars{1 + a}/2}} - {a \over 2}\sum_{k = 0}^{\infty} {1 \over \pars{k + \pars{1 - a/2}}\pars{k + \pars{1 + a/2}}} \\[3mm]&= {1 \over a} + {a \over 2}\,{\Psi\pars{\bracks{1 - a}/2} - \Psi\pars{\bracks{1 + a}/2} \over \pars{1 - a}/2 - \pars{1 + a}/2} - {a \over 2}\,{\Psi\pars{1 - a/2} - \Psi\pars{1 + a/2} \over \pars{1 - a/2} - \pars{1 + a/2}} \\[3mm]&= {1 \over a} + \half\bracks{ -\Psi\pars{1 - a \over 2} + \Psi\pars{1 + a \over 2} + \Psi\pars{1 -{a \over 2}} -\Psi\pars{1 + {a \over 2}}} \end{align}

With the identities: $$\Psi\pars{1 - z} = \Psi\pars{z} + \pi\cot\pars{\pi z}\,,\quad \Psi\pars{\half + z} = \Psi\pars{\half - z} + \pi\tan\pars{\pi z}\,,\quad \Psi\pars{1 + z} = \Psi\pars{z} + {1 \over z}$$ we'll get

\begin{align} &\color{#0000ff}{\large \sum_{k = - \infty }^\infty{\pars{-1}^{k} \over a + k}} = {1 \over a} + \half\bracks{ \pi\tan\pars{\pi a \over 2} + \Psi\pars{a \over 2} + \pi\cot\pars{\pi a \over 2} - \Psi\pars{a \over 2} - {2 \over a}} \\[3mm]&= {\pi \over 2}\,{\sec^{2}\pars{\pi a/2} \over \tan\pars{\pi a/2}} = {\pi \over \sin\pars{\pi a}} = \color{#0000ff}{\large\pi\,\csc\pars{\pi a}} \end{align}
-
You seem to be forgetting the $k=0$ term. The result should be $\pi\csc \pi a$ as seen here. – Pedro Tamaroff Dec 2 '13 at 1:38
@PedroTamaroff Yes. I already checked. I forgot to add the $k = 0$ term and the $1/2$ factor. Thanks. – Felix Marin Dec 2 '13 at 2:09

Integrate over a closed contour that is in the shape of a wedge in the 1st quadrant, having an angle of $\pi/5$. That is, consider

$$\oint_C dz \frac{z^2}{z^{10}+1}$$

where $C$ is that wedge contour. As mentioned above, the contour splits into 3 pieces:

$$\oint_C dz \frac{z^2}{z^{10}+1} = \int_0^R dx \frac{x^2}{x^{10}+1} + i R \int_0^{\pi/5}d\phi \, e^{i \phi} \frac{R^2 e^{i 2 \phi}}{R^{10} e^{i 10 \phi}+1}+ e^{i 3 \pi/5} \int_R^0 dx \frac{x^2}{x^{10}+1}$$

As $R \to \infty$, the second integral vanishes as $1/R^7$ (Note that the denominator never vanishes). The rest is then just equal to $i 2 \pi$ times the residue at the pole $z=e^{i \pi/10}$. Thus, combining the integrals, we have

$$\left (1-e^{i 3 \pi/5}\right) \int_0^{\infty} dx \frac{x^2}{x^{10}+1} = i 2 \pi \frac{e^{i 2 \pi/10}}{10 e^{i 9 \pi/10}}$$

Doing the algebra, I get

$$\int_0^{\infty} dx \frac{x^2}{x^{10}+1} = \frac{\pi}{10 \cos{(\pi/5)}}$$

-
This seems similar to Robert Israel's completed answer. – robjohn May 15 '13 at 11:24
Were you answering this question? – robjohn May 15 '13 at 11:44
@robjohn: yes. As I had invested the time to do so, I placed the answer in the non-dup section. – Ron Gordon May 15 '13 at 11:59
I don't understand why you have answered with a particular case. I also don't know any complex integration, so I cannot make use of this, yet. – Pedro Tamaroff May 15 '13 at 13:18
@PeterTamaroff: The original question was closed. due to being a duplicate of this; it was closed just as I was finishing typing this solution. – Ron Gordon May 15 '13 at 14:35