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.