Help with a Cauchy Integral

Eliot Rosewater

I'm stuck again, and with the same subject! This time it's a Cauchy Integral. I nearly broke my hand in frustration yesterday.

The problem is to find the value of

[LATEX]\displaystyle I= \int_{-\infty}^{\infty}\frac{\cos(\pi x)}{x^2-2x+2}dx[/LATEX]

(The answer is [LATEX]\displaystyle -\frac{\pi}{e^{\pi}}[/LATEX])

So I'm doing using the Cauchy Residue Theorem. I first define the function as a complex function and factorize the bottom:

[LATEX]\displaystyle f(z) = \frac{\cos(\pi z)}{z^2-2z+2}[/LATEX]
[LATEX]\displaystyle f(z) = \frac{\cos(\pi z)}{(z-(1+i))(z-(1-i))}[/LATEX]

So there are two poles of order 1: one at 1+i and one at 1-i.

I use a semi-circular contour in the upper plane, something like this:



The straight line part is I, and the circular part is, say, C. Now, as the radius of the semi-circle goes to infinity the value of C goes to 0. So, using the Cauchy Residue Theorem, we have

[LATEX]\displaystyle I = 2 \pi i \text{Res}[f(z); z=1+i][/LATEX]

I now use the residue formula:

[LATEX]\displaystyle \text{Res}[f(z); z=1+i] = (z-(1+i))f(z)[/LATEX]

[LATEX]\displaystyle \text{Res}[f(z); z=1+i] = (z-(1+i))\frac{\cos(\pi z)}{(z-(1+i))(z-(1-i))}[/LATEX]

[LATEX]\displaystyle \text{Res}[f(z); z=1+i] = \frac{\cos(\pi z)}{(z-(1-i))}[/LATEX]

Let z = 1+i.

[LATEX]\displaystyle \text{Res}[f(z); z=1+i] = \frac{\cos(\pi (1+i))}{((1+i)-(1-i))}[/LATEX]

[LATEX]\displaystyle \text{Res}[f(z); z=1+i] = \frac{\cosh(\pi)}{(2i)}[/LATEX]

And so:

[LATEX]\displaystyle I = 2\pi i \frac{\cosh(\pi)}{(2i)}[/LATEX]

[LATEX]\displaystyle I = \pi \cosh(\pi)[/LATEX]

Which isn't the right answer. Can anyone see where I'm going wrong?

Fremen

Double check it using a partial fraction expansion, maybe?

Fremen

Wait, can you really use Jordan's lemma here? Doesn't Cos blow up on the imaginary axis?

Fremen

One strategy here might be to solve the integral for e^(i pi x) in place of Cos in the numerator. Then maybe you can use e^(i pi x) = cos(pi x) + i sin(pi x) and equate real and imaginary parts.

Eliot Rosewater

Fremen wrote: »

Wait, can you really use Jordan's lemma here? Doesn't Cos blow up on the imaginary axis?

Hmm, I think so now. I did an empirical test on Mathematica, and its absolute value seemed bounded above by 2, but on mature reflection I think that's rubbish. Stupid empiricism!

I did it your way with the exponentials and I got a funny answer, I'll try that again though.

Thanks for the help. The notes I'm working off don't even mention Jordan's lemma.

Fremen

I guess you're not used to thinking of cos as a sum of complex exponentials.

When you write

cos(z) = (e^iz + e^-iz)/2 it's pretty clear it blows up along the line z = ci.

The result is almost the same as your derivation but when you evaluate the residues, instead of a cos(pi*(1+i))/2i, you have an e^(i*pi*(1+i))/2i = (e^(i*pi) e^(-pi))/2i.

Then you hit it with the factor of 2*pi*i and your answer pops out. The sin component is odd so it integrates to 0.

Jordan's lemma is fundamental here - it's the reason e^i pi z works but cos(pi z) doesn't. It would be very strange to discuss this stuff without mentioning it. There's a derivation of a very similar result on the wikipedia page.

Eliot Rosewater

That's great Fremen - it's all sorted now! Thanks for your help and patience!