Integration Over a Line in the Complex Plane

[Euge]

For $c > 0$ and $0 \le x \le 1$, find the complex integral $$\int_{c - \infty i}^{c + \infty i} \frac{x^s}{s}\, ds$$

 

[anuttarasammyak]

Spoiler

Putting s=c+iu, The integral is
\int_{-\infty}^{+\infty}\frac {e^{(c+ui)\ln x}} { c+ui } i du
=e^{c\ln x}\int_{-\infty}^{+\infty}\frac{e^{ui\ln x}}{u-ic}du
Applying residue theorem for complex integral of u
=0

 

[pasmith]

Spoiler

Set x = e^{t}. Then we have <br /> \int_{c-i\infty}^{c+i\infty} \frac{x^s}{s}\,ds = <br /> \int_{c-i\infty}^{c+i\infty} \frac{e^{st}}{s}\,ds = f(t). But comparison to the inverse Laplace transform of 1/s shows that <br /> f(t) = 2\pi i U(t) and since t = \ln x \leq 0 we have f(t) = 0.

 

[Euge]

Thanks for participating! However, the solutions are incorrect. One has to take more care when integrating over a complex line.

 

[Euge]

Note: For at least one value of $x$ the line integral is nonzero.

 

[Paul Colby]

Well, I get the same result. I expect there is a misprint in the problem statement or I too have made a sign error.

Spoiler

Make the change of variables $s = c + iz$ which makes the integration path the real $z$ axis.

$$\frac{x^s}{s}ds = i\frac{e^{(c+iz)\ln{x}}}{c+iz}dz = x^{c} \frac{e^{i(\ln{x})z}}{z-ic}dz$$

This puts the pole in $z$ at $ic$ which is in the upper complex plane. Normally in problems like these one closes the contour with a large semi-circle which encloses the pole. Clearly $\ln{x} < 0$ which means

$$e^{i(\ln{x})z}$$

grows without bound as the imaginary part of $z$ becomes large. This forces us to close the contour in the negative imaginary direction excluding the pole. Like both the other contributors, I get 0 for the value when $0\le x < 1$.

 

[Euge]

Hi @Paul Colby, you missed the $x = 1$ case!

 

[Paul Colby]

Hi @Paul Colby, you missed the $x = 1$ case!

I excluded it intensionally.

[edit] Actually, this brings up a somewhat philosophical issue. In physics (at least the stuff I've done) one deals with analytic functions. By all rights, the above continues to 0 everywhere in the complex plane. The set of measure 0 stuff, again from the point of view of doing physics, isn't all that interesting. Now, you say, hold on one minute. The step function (like from the Laplace transform approach), is certainly of interest in physical problems. And, you'd be right. But the Gibbs phenomena stuff, eh, not so much.

 

[Euge]

Spoiler

If $x = 1$, then the integral is $$\lim_{R\to \infty} \int_{c - Ri}^{c + Ri} \frac{ds}{s} = \lim_{R\to \infty} \int_{-R}^R \frac{i\, dt}{c + it} = \lim_{R\to \infty} \int_{-R}^R \frac{t + ci}{c^2 + t^2}\, dt = \lim_{R\to \infty} i\int_{-R}^R \frac{dt}{c^2 + t^2}$$which evaluates to $$\lim_{R\to \infty} 2i \arctan\left(\frac{R}{c}\right) = \pi i$$