Integral of x^m/(x^n+a^n)^p: Contour Integration

[captain]

can anyone show me step by step of how to evaluate the integral of
[x^m/(x^n+a^n)^p](dx) from negative infinity to positive infinity. all i know is that contour integration is required to solve this problem.

 

[AiRAVATA]

You'll have to use Residue Theorem. In order to do that, consider the complex integral

\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=\int_{\gamma_1}\frac{z^m}{(z^n+a^n)^p}dz +\int_{\gamma_{2}}\frac{z^m}{(z^n+a^n)^p}dz,

where \gamma_1=(x,0) and \gamma_2=Re^{i\theta}, x\in(-R,R) and \theta\in(0,\pi) (i.e. \gamma is the semicircle of radius R enclosing the upper semiplamen \mathbb{H}^+). Then

\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=\int_{-R}^R \frac{x^m}{(x^n+a^n)^p}dx+\int_0^\pi \frac{R^m e^{im\theta}}{(R^n e^{in\theta}+a^n)^p}iRe^{i\theta}d\theta.

Now you have to prove that as R\rightarrow \infty the second integral goes to zero, so

\left|\int_0^\pi \frac{R^m e^{im\theta}}{(R^n e^{in\theta}+a^n)^p}iRe^{i\theta}d\theta\right|\le \frac{\pi R^{m+1}}{(R^n-a^n)^p}, \qquad R\gg |a|

and therefore m,n and p cannot be arbitrary (doh!). So assuming m+1<np, then

\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=\int_{-\infty}^\infty \frac{x^m}{(x^n+a^n)^p}dx.

Now, using the Residue Theorem

\int_{-\infty}^\infty \frac{x^m}{(x^n+a^n)^p}dx=\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=2\pi i \sum_i Res\left(\frac{z^m}{(z^n+a^n)^p},a_i\right),

where the a_i are roots of the equation z^n+a^n=0 in \mathbb{H}^+.

It's all straightforward from here.---EDIT---

I'm also assuming that m,n,p \in \mathbb{N}. If not, then you'll have to pick branches of the logarithm and the problem is a lot different.

 

[Kummer]

AIRAVATA did a good job showing the remainder goes to zero. This happens to be a special case of a more general case.

Theorem: Consider a complex function f(z) having the following properties:
1)f is meromorphic in upper half plane having finitely many poles (c_j)\ 1\leq j\leq n.
2)c_j\not \in \mathbb{R}, i.e. not on real line.
3)\lim_{|z|\to \infty}zf(z) = 0 with z in upper half plane.
Then,
\mbox{PV}\int_{-\infty}^{\infty} f(x) dx = 2\pi i \sum_{j=1}^n \mbox{res}(f,c_j)

EDIT: Not if m,n,p\not \in \mathbb{N} then the theorem does not apply since it is no longer meromorphic.

 

[captain]

You'll have to use Residue Theorem. In order to do that, consider the complex integral

\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=\int_{\gamma_1}\frac{z^m}{(z^n+a^n)^p}dz +\int_{\gamma_{2}}\frac{z^m}{(z^n+a^n)^p}dz,

where \gamma_1=(x,0) and \gamma_2=Re^{i\theta}, x\in(-R,R) and \theta\in(0,\pi) (i.e. \gamma is the semicircle of radius R enclosing the upper semiplamen \mathbb{H}^+). Then

\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=\int_{-R}^R \frac{x^m}{(x^n+a^n)^p}dx+\int_0^\pi \frac{R^m e^{im\theta}}{(R^n e^{in\theta}+a^n)^p}iRe^{i\theta}d\theta.

Now you have to prove that as R\rightarrow \infty the second integral goes to zero, so

\left|\int_0^\pi \frac{R^m e^{im\theta}}{(R^n e^{in\theta}+a^n)^p}iRe^{i\theta}d\theta\right|\le \frac{\pi R^{m+1}}{(R^n-a^n)^p}, \qquad R\gg |a|

and therefore m,n and p cannot be arbitrary (doh!). So assuming m+1<np, then

\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=\int_{-\infty}^\infty \frac{x^m}{(x^n+a^n)^p}dx.

Now, using the Residue Theorem

\int_{-\infty}^\infty \frac{x^m}{(x^n+a^n)^p}dx=\int_\gamma \frac{z^m}{(z^n+a^n)^p}dz=2\pi i \sum_i Res\left(\frac{z^m}{(z^n+a^n)^p},a_i\right),

where the a_i are roots of the equation z^n+a^n=0 in \mathbb{H}^+.

It's all straightforward from here.


---EDIT---

I'm also assuming that m,n,p \in \mathbb{N}. If not, then you'll have to pick branches of the logarithm and the problem is a lot different.

how would you arrive at your final answer. i understand how you have done it so far but i have no idea how arrive at the final answer which has the gamma function and sin in it. I actually have the final answer right here: http://www.sosmath.com/tables/integral/integ41/integ41.html it is the last integral at the bottom of the page (or #8).

 

[AiRAVATA]

Well, I can assure you is not an easy task but it is possible. Although it migh have nothing to do with what I posted before.

I present to you the beta function:

B(\xi,\eta)=\int_0^\infty \frac{u^{\xi-1}}{(1+u)^{\xi+\eta}}du,\qquad \hbox{Re }\xi>0,\quad \hbox{Re }\eta>0,

wich can be written as*

B(\xi,\eta)=\frac{\Gamma(\xi)\Gamma(\eta)}{\Gamma(\xi+\eta)}.

Also, the Gamma function satisfies the following identity*

\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z}.

So, taking the change of variable x^n/a^n[/tex] in the Beta function, we have<br /> <br /> B(\xi,\eta)=na^{n\eta}\int_0^\infty \frac{x^{n\xi-1}}{(a^n+x^n)^{\xi+\eta}}dx.<br /> <br /> Then, taking \xi=(m+1)/n and \eta=p-(m+1)/n, you have the following relation between your integral and the Beta function:<br /> <br /> \int_0^\infty \frac{x^m}{(a^n+x^n)^p}dx=\frac{a^{m+1-np}}{n}B\left(\frac{m+1}{n}, p-\frac{m+1}{n}\right),<br /> <br /> and I assume is straight sustitution and Gamma function manipulation from here to the answer you want.<br /> <br /> *The proof of this relations can be found in<br /> N.N. LEBEDEV, <i>Special Functions and their Applications</i>, Dover Publications Inc, New York, 1972.

 

[captain]

thaks man i was having trouble with that problem for quite a while now

 

[Kummer]

The identity can also be established by computing,
\int_{-\infty}^{\infty} \frac{e^{ax}}{e^x+1} dx on the rectangle \pm R \pm \pi i, \pm R. And then using a Beta substitution this converts into the Gamma integral.

 

[captain]

Well, I can assure you is not an easy task but it is possible. Although it migh have nothing to do with what I posted before.

I present to you the beta function:

B(\xi,\eta)=\int_0^\infty \frac{u^{\xi-1}}{(1+u)^{\xi+\eta}}du,\qquad \hbox{Re }\xi&gt;0,\quad \hbox{Re }\eta&gt;0,

wich can be written as*

B(\xi,\eta)=\frac{\Gamma(\xi)\Gamma(\eta)}{\Gamma(\xi+\eta)}.

Also, the Gamma function satisfies the following identity*

\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z}.

So, taking the change of variable x^n/a^n[/tex] in the Beta function, we have<br /> <br /> B(\xi,\eta)=na^{n\eta}\int_0^\infty \frac{x^{n\xi-1}}{(a^n+x^n)^{\xi+\eta}}dx.<br /> <br /> Then, taking \xi=(m+1)/n and \eta=p-(m+1)/n, you have the following relation between your integral and the Beta function:<br /> <br /> \int_0^\infty \frac{x^m}{(a^n+x^n)^p}dx=\frac{a^{m+1-np}}{n}B\left(\frac{m+1}{n}, p-\frac{m+1}{n}\right),<br /> <br /> and I assume is straight sustitution and Gamma function manipulation from here to the answer you want.<br /> <br /> *The proof of this relations can be found in<br /> N.N. LEBEDEV, <i>Special Functions and their Applications</i>, Dover Publications Inc, New York, 1972.

<br /> <br /> <br /> do you have link to access that book as a pdf file?

 

[AiRAVATA]

Nope, I have the book right here in my right. There must be one in your local library. There is a lot of information regarding the Gamma and Beta functions in the wikipedia. I'm sure it will be of some help.