Application of complex analysis to real integrals

[jimlyn]

Homework Statement

solve integral x^3/(e^x-1) with limits from 0 to infinity

Homework Equations

The Attempt at a Solution

i tried using a rectangular contour,the boundaries of the contour pass through z=0 but the complex equivalent has pole at z=0. by Cauchy theorem the function must be analytic even at the boundaries.how do i tackle this problem

 

[Amok]

Wait, are you sure that's the correct integral? It has no singularities and no poles. Can you be more specific about what you are trying to do?

 

[Petar Mali]

If I remember well this integral is from Stefan - Boltzman law. It's value is \frac{\pi^4}{15}. Put e^x in form of the series, use some approximation and you must in the and get Riman zeta function.

 

[Gib Z]

Amok, it actually has an infinite number of simple poles. You may want to recheck your thinking.

jimlyn - There is no pole at z=0. There is a removable singularity, and using in the theory of analytic functions, if there is ever a removable singularity we assume it is already redefined such that it is now analytic at that point.

Nevertheless, if you want to dodge a pole that's blocking your path, just go around it. Indent your path with a little semi-circle going around the pole, and take the limit as the radius of that semi circle tends to zero.

We can avoid contour integration all together, by noting 0 < e^{-x} < 1 in (0,\infty), so we can expand as a geometric series like this:

\int^{\infty}_0 \frac{x^3}{e^x-1} dx = \int^{\infty}_0 \frac{x^3 e^{-x} }{1-e^{-x} } dx = \int^{\infty}_0 \left( x^3 e^{-x} + x^3 e^{-2x} + x^3 e^{-3x} + \cdots \right) dx

We can split up the integral and integrate each term individually (justified by Lebesgue's Monotone convergence theorem). Hopefully the rest is simple.

 

[Amok]

Amok, it actually has an infinite number of simple poles. You may want to recheck your thinking.

jimlyn - There is no pole at z=0. There is a removable singularity, and using in the theory of analytic functions, if there is ever a removable singularity we assume it is already redefined such that it is now analytic at that point.

Nevertheless, if you want to dodge a pole that's blocking your path, just go around it. Indent your path with a little semi-circle going around the pole, and take the limit as the radius of that semi circle tends to zero.

We can avoid contour integration all together, by noting 0 < e^{-x} < 1 in (0,\infty), so we can expand as a geometric series like this:

\int^{\infty}_0 \frac{x^3}{e^x-1} dx = \int^{\infty}_0 \frac{x^3 e^{-x} }{1-e^{-x} } dx = \int^{\infty}_0 \left( x^3 e^{-x} + x^3 e^{-2x} + x^3 e^{-3x} + \cdots \right) dx

We can split up the integral and integrate each term individually (justified by Lebesgue's Monotone convergence theorem). Hopefully the rest is simple.

I didn't really understand what he meant, I read the integral as:

\int^{\infty}_0 \frac{x^3}{e^{x-1}} dx

My bad.