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:

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

which 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 [itex]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.[/itex]

 

[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:

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

which 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 [itex]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.[/itex]

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

 

[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.