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

In summary, the conversation discusses how to evaluate the integral of [x^m/(x^n+a^n)^p](dx) from negative infinity to positive infinity. The solution involves using contour integration and the Residue Theorem, and then using the Beta function and Gamma function to arrive at the final answer.
  • #1
captain
164
0
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.
 
Physics news on Phys.org
  • #2
You'll have to use Residue Theorem. In order to do that, consider the complex integral

[tex]\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,[/tex]

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

[tex]\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.[/tex]

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

[tex]\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| [/tex]

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

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

Now, using the Residue Theorem

[tex]\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),[/tex]

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

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

I'm also assuming that [itex]m,n,p \in \mathbb{N}[/itex]. If not, then you'll have to pick branches of the logarithm and the problem is a lot different.
 
Last edited:
  • #3
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 [tex]f(z)[/tex] having the following properties:
1)[tex]f[/tex] is meromorphic in upper half plane having finitely many poles [tex](c_j)\ 1\leq j\leq n[/tex].
2)[tex]c_j\not \in \mathbb{R}[/tex], i.e. not on real line.
3)[tex]\lim_{|z|\to \infty}zf(z) = 0[/tex] with [tex]z[/tex] in upper half plane.
Then,
[tex]\mbox{PV}\int_{-\infty}^{\infty} f(x) dx = 2\pi i \sum_{j=1}^n \mbox{res}(f,c_j)[/tex]

EDIT: Not if [tex]m,n,p\not \in \mathbb{N}[/tex] then the theorem does not apply since it is no longer meromorphic.
 
  • #4
AiRAVATA said:
You'll have to use Residue Theorem. In order to do that, consider the complex integral

[tex]\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,[/tex]

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

[tex]\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.[/tex]

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

[tex]\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| [/tex]

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

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

Now, using the Residue Theorem

[tex]\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),[/tex]

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

It's all straightforward from here.


---EDIT---

I'm also assuming that [itex]m,n,p \in \mathbb{N}[/itex]. 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).
 
  • #5
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:

[tex]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,[/tex]

wich can be written as*

[tex]B(\xi,\eta)=\frac{\Gamma(\xi)\Gamma(\eta)}{\Gamma(\xi+\eta)}.[/tex]

Also, the Gamma function satisfies the following identity*

[tex]\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z}.[/tex]

So, taking the change of variable [itex]x^n/a^n[/tex] in the Beta function, we have

[tex]B(\xi,\eta)=na^{n\eta}\int_0^\infty \frac{x^{n\xi-1}}{(a^n+x^n)^{\xi+\eta}}dx.[/tex]

Then, taking [itex]\xi=(m+1)/n[/itex] and [itex]\eta=p-(m+1)/n[/itex], you have the following relation between your integral and the Beta function:

[tex]\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),[/tex]

and I assume is straight sustitution and Gamma function manipulation from here to the answer you want.

*The proof of this relations can be found in
N.N. LEBEDEV, Special Functions and their Applications, Dover Publications Inc, New York, 1972.
 
Last edited:
  • #6
thaks man i was having trouble with that problem for quite a while now
 
  • #7
The identity can also be established by computing,
[tex]\int_{-\infty}^{\infty} \frac{e^{ax}}{e^x+1} dx[/tex] on the rectangle [tex]\pm R \pm \pi i, \pm R[/tex]. And then using a Beta substitution this converts into the Gamma integral.
 
  • #8
AiRAVATA said:
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:

[tex]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,[/tex]

wich can be written as*

[tex]B(\xi,\eta)=\frac{\Gamma(\xi)\Gamma(\eta)}{\Gamma(\xi+\eta)}.[/tex]

Also, the Gamma function satisfies the following identity*

[tex]\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin \pi z}.[/tex]

So, taking the change of variable [itex]x^n/a^n[/tex] in the Beta function, we have

[tex]B(\xi,\eta)=na^{n\eta}\int_0^\infty \frac{x^{n\xi-1}}{(a^n+x^n)^{\xi+\eta}}dx.[/tex]

Then, taking [itex]\xi=(m+1)/n[/itex] and [itex]\eta=p-(m+1)/n[/itex], you have the following relation between your integral and the Beta function:

[tex]\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),[/tex]

and I assume is straight sustitution and Gamma function manipulation from here to the answer you want.

*The proof of this relations can be found in
N.N. LEBEDEV, Special Functions and their Applications, Dover Publications Inc, New York, 1972.


do you have link to access that book as a pdf file?
 
  • #9
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.
 

1. What is the purpose of using contour integration when calculating the integral of x^m/(x^n+a^n)^p?

Contour integration is a powerful technique used to evaluate complex integrals that cannot be easily solved by standard integration methods. It involves transforming the original integral into a contour integral, which can then be evaluated by utilizing the properties of complex functions.

2. How do I choose the appropriate contour when using contour integration for this integral?

The choice of contour depends on the behavior of the integrand and the location of the poles. Generally, a contour that avoids all singularities and encloses the region of interest is a good choice. It is also important to consider the orientation and direction of the contour, which can affect the final result.

3. What are the key steps involved in solving the integral of x^m/(x^n+a^n)^p using contour integration?

The key steps involved in solving this integral include:

  1. Identifying the appropriate contour
  2. Breaking up the integrand into partial fractions, if necessary
  3. Utilizing the Cauchy Residue Theorem to evaluate the integral
  4. Applying the limit as the radius of the contour goes to infinity to eliminate any contributions from the contour's arc

4. Are there any special cases that I need to consider when using contour integration for this integral?

Yes, there are a few special cases that may arise when using contour integration for this integral. These include:

  • Multiple poles located on the contour
  • Poles located on the contour's arc
  • Branch cuts in the integrand
In these cases, additional techniques such as the Residue at Infinity or the Residue at Branch Points may need to be applied.

5. Can contour integration be used to solve all integrals involving complex functions?

No, contour integration is not a universally applicable method for solving all integrals involving complex functions. It is most useful when dealing with integrals that have singularities or branch cuts, and may not be suitable for other types of integrals. Other integration techniques, such as substitution or partial fractions, may be more appropriate in certain cases.

Similar threads

Replies
16
Views
2K
  • Calculus
Replies
1
Views
963
Replies
1
Views
830
  • Calculus
Replies
8
Views
2K
Replies
3
Views
1K
Replies
31
Views
749
Replies
2
Views
1K
Replies
14
Views
1K
Back
Top