# Complex Integral

1. Nov 25, 2005

I attempted to prove the following equality, but to no avail. Anyone is willing to lend a hand?

$$\int_0^{\infty} s^{2t-2v} e^{i w s} ds + \int_0^{\infty} s^{2t-2v} e^{-i w s} ds = \left[ \left( \frac{1}{-iw}\right)^{2t-2v+1} + \left( \frac{1}{iw}\right)^{2t-2v+1} \right] \Gamma(2t-2v + 1)$$,

where $$i = \sqrt{-1}$$, $$s > 0$$, $$\Gamma(\cdot)$$ is gamma function, $$-\pi \leq w \leq \pi$$, $$0 < t <1$$, and $$1 \leq v \leq \infty$$ is integer.

I got almost all the RHS, except the power terms. It seems strange as IMHO it is only true when $$t$$ is integer and that $$2t-2v+1 \geq 0$$.

Last edited: Nov 25, 2005
2. Nov 25, 2005

### benorin

Try this.

In the first integral, apply this substitution:

Put $u=-iws$ (to make the exponent of e $-u$ so that the integral will look more like the gamma function), which gives $s=\frac{u}{-i w}$; hence $ds=\frac{du}{-iw}$, and the integral becomes

$$\int_0^{\infty} s^{2t-2v} e^{iws} ds = \int_0^{\infty} \left( \frac{u}{-i w}\right) ^{2t-2v} e^{-u} \frac{du}{-iw}= \left( \frac{1}{-iw}\right) ^{2t-2v+1}\int_0^{\infty}u^{2t-2v} e^{-u} du = \left( \frac{1}{-iw}\right) ^{2t-2v+1} \Gamma\left( 2t-2u+1\right)$$

The second integral can be evaluated by a similar substitution, try it.

Last edited: Nov 25, 2005
3. Nov 25, 2005

Gosh! :surprised Is it that simple? I did integration by parts that was 3 pages long and yet got nothing!

I have one question though. Since $$u = -iws$$, so $$u = -i\infty$$ for $$s = \infty$$ and $$u = 0$$ for $$s = 0$$. Wouldn't it be

$$\left( \frac{1}{-iw}\right)^{2t-2v+1} \int_0^{-i\infty}u^{2t-2v} e^{-u} du$$

My question is, will

$$\int_0^{-i\infty}u^{2t-2v} e^{-u} du = \Gamma\left( 2t-2v+1\right)$$

valid?

p/s: I am assuming $$w > 0$$ for the sake of this discussion.

Last edited: Nov 25, 2005
4. Nov 25, 2005

### benorin

I have asked this question before, and yes, it is legitimate. In the extended complex plane, the point at infinity is the common "terminal point" of any ray eminating from the origin, or, if you will, the north pole of the Riemann sphere. So, in the common English: "infinity is infinity."

5. Nov 25, 2005

### shmoe

The contours may have the same "infinity" endpoint in the extended complex plane, but that doesn't mean you can change contours at a whim. Consider $\int_0^\infty e^{-u}du$ and change the contour of integration to $\int_0^{i \infty} e^{-u}du$, the first integral converges, the second does not.

This question is a little strange, you need $$0> 2t-2v> -1$$ for convergence- the upper bound needed as you approach infinity, the lower bound as you approach 0. So I'm not sure about their allowed values of v and t? Anyways, if you satisfy this inequality then you can show this change of contours is permissible by considering the contour with four pieces, the line segment from r to R, the quarter circle from R to iR with center 0, the line from iR to ir, and finally the quarter circle from ir to r with center 0. Show that as r->0 and R->infinity (independantly) that the integrals over these semicircles both vanish. (also consider any possible residues inside this contour!)

Or you can look this up in a table. You can combine the two integrals and look for the Mellin transform of cos (which can be derived by rotating the contour to the imaginary axis as above).

Last edited: Nov 25, 2005
6. Nov 27, 2005

### benorin

Thank you shmoe.