Can Cauchy's Theorem Help Evaluate Integrals with Poles?

[zetafunction]

i want to perform the following integrals

\int_{-\infty}^{\infty}dx \frac{f(x)}{x^{2}-a^{2}}

the problem is that the integral has poles at x=a and x=-a , could we apply

i think this is the definition of Hadamard finite part integral, performing an integral with singularities by means of Cauchy's theorem.

 

[Count Iblis]

You can compute the Principal Part of the Integral. If you can apply the Residue therorem, then you find that the residues at x = a and x = -a count for half. This is known as Plemelj's formula.

 

[HallsofIvy]

Cauchy's theorem only applies to integrals around a closed path in the complex plane. That path also cannot include any singularities of the function. You could do this by integrating along the real axis, from -R to R, with small half circles, radius \epsilon around -a and a in the upper half plane, then by a half circle from R to -R. Since the function is analytic inside that path, Cauchy's theorem gives 0 for the integral around the entire path. I think it would be easy to show that the limit for the upper half circle, as R goes to infinity, is 0. Thus the problem reduces to determining the integral around those small half circles around -a and a.

 

[zetafunction]

yes of course but since there are real poles at x=a and x=-a then the integral would be infinite , this is a singular integral, from the beginnig and this can not be 'regularized' to give a finite real value isn't it ??

 

[squidsoft]

I think it depends on what f(x) is. For example, compute:

P.V.\int_{-\infty}^{\infty} \frac{f(x)}{x^2-a^2}dx

for:

f(x)=x,x^2,x^3,\frac{1}{x-i}

 

[zetafunction]

nope due to the factor 2\pi i the integral will be an imaginary number.

 

[Count Iblis]

nope due to the factor 2\pi i the integral will be an imaginary number.

Which is reasonable. As you move from one side of the 1/(x-a) and
1/(x+a) singularities at x = a and -a to the other side, the argument of the integrals in the neighborhood of the singularities ( Log(x+a) and Log(x-a) ) pick up a factor exp(i pi).

 

[squidsoft]

Hi. Here's what I calculated for two integrals:

P.V. \int_{-\infty}^{\infty}\frac{x}{(x-1)(x+1)}dx=\pi i-\pi i=0

and:

P.V. \int_{-\infty}^{\infty}\frac{1}{(x-i)(x-1)(x+1)}dx=-\pi i+\pi i(1/2)=-\pi i/2

 

[Mute]

nope due to the factor 2\pi i the integral will be an imaginary number.

If f(x) is a real function then your integral will be real. Any 2\pi i terms that appear will necessarily be eliminated, either by some other factor of i appearing to cancel that i or the integral being zero.

Since the poles of the integral are simple poles, you can apply the half residue theorem, and so

P.V.\int_{-\infty}^{\infty}dx~\frac{f(x)}{(x-a)(x+a)} = \pi i\left[f(-a) + f(a)\right]

AS LONG AS f(z), where z = x + iy, is analytic within the upper half plane, AND AS LONG AS |f(Re^{i\theta})| < R as R \rightarrow \infty, otherwise the contribution from the large arc will not go to zero. Note that the analyticity condition must be taken into account for the second integral squidsoft did, which has a pole at z = i, which is inside the contour. Accordingly, looking at squidsoft's results,


P.V. \int_{-\infty}^{\infty}\frac{x}{(x-1)(x+1)}dx=\pi i-\pi i=0

is correct, but

P.V. \int_{-\infty}^{\infty}\frac{1}{(x-i)(x-1)(x+1)}dx=-\pi i+\pi i(1/2)=-\pi i/2

is not quite what I seem to get. I find

\int_\gamma dz~\frac{1}{(z-i)(z-1)(z+1)} = \frac{2\pi i}{((i)^2-1)} = P.V.\int_{-\infty}^{\infty}dx~\frac{1}{(x-i)(x-1)(x+1)} - \pi i \left[\frac{1}{2(1-i)} + \frac{1}{2(1+i)} \right]

where the "half residue terms" on the RHS came from the small arcs, traversed clockwise, hence the minus sign. Solving,

P.V.\int_{-\infty}^{\infty}dx~\frac{1}{(x-i)(x-1)(x+1)} = -\frac{3\pi i}{4}

The discrepancy appears to be accidentally dropping a factor of 1/2 on the half residue terms.