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I've come across the following derivation, for example here, for the following Laplace transform of a power:

$$\mathcal L \{ t^a \} = \frac {\Gamma(a+1)}{s^{a+1}}$$

For $\operatorname{Re}(a) > -1, \operatorname{Re}(s)>0$. The proof I've seen does the following:

$$\mathcal L\{t^a\} := \int_0^{\to +\infty}t^a e^{-st} \, \mathrm dt$$

$$ = s^{-1} \int_0^? \left({\frac u s}\right)^a e^{-u}\, \mathrm du$$

and so on.

The upperbound of the integral is what I'm questioning. The proofs I've seen have $\infty$ in the upper bound, but I don't see how you can replace the complex number $st$ with a real number $u$ without making the bounds of the integral complex. But that would turn the Riemann integral into a contour integral! How is that allowed? Is the proof wrong?

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GFauxPas
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  • As Dr.MV said you need some complex-analysis theorems for showing that $\int_0^\infty t^a e^{-st}dt = s^{-a-1} \int_0^\infty t^a e^{-t}dt= s^{-a-1} \Gamma(a+1)$ (whenever both converge, that is $Re(s) > 0,Re(a) > -1$) – reuns Nov 20 '16 at 00:36

1 Answers1

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Let $s=\sigma+i\omega \in \mathbb{C}$, where $\sigma>0$ and $\omega \in \mathbb{R}$.

Let $I(a,s)$ be given by the integral

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty t^ae^{-st}\,dt=\lim_{L\to \infty}\int_0^L t^ae^{-st}\,dt}\tag 1$$

Now, enforcing the substitution $z=st$ in the integral on the right-hand side of $(1)$ reveals

$$\int_0^L t^ae^{-st}\,dt=\frac{1}{s^{a+1}}\int_0^{L\sigma+iL\omega}z^ae^{-z}\,dz \tag2$$

Next, we move to the complex plane. Let $f(z)=z^ae^{-z}$. If $a$ is not an integer, then $z^a$ is multivalued and has a branch point at $z=0$. If we cut the plane along the negative real axis, then $z^a$ is holomorphic on $\mathbb{C}\setminus \{z|\text{Re}(z)\le 0, \text{Im}(z)=0\}$.

Using Cauchy's Integral Theorem, we have

$$\oint_C z^ae^{-z}\,dz =0 \tag 3$$

where $C$ is a closed contour, comprised of $(i)$ the line segment from $0$ to $L\sigma$, $(ii)$ the line segment from $L\sigma$ to $L\sigma+iL\omega$, and $(iii)$ the line segment from $L\sigma+iL\omega$ to $0$. (We also need to deform the contour around the branch point, but the contribution to the integral can be made arbitrarily small).

Therefore, we can write $(3)$ as

$$\int_0^{L\sigma}z^ae^{-z}\,dz+\int_{L\sigma}^{L\sigma+iL\omega}z^ae^{-z}\,dz+\int_{L\sigma+iL\omega}^0 z^ae^{-z}\,dz=0 \tag 4$$

As $L\to \infty$, the second integral on the left-hand side of $(4)$ approaches zero. Therefore, letting $L\to \infty$ in $(4)$ yields

$$\int_0^{\infty}t^ae^{-t}\,dt=\lim_{L\to \infty}\int_0^{L(\sigma+i\omega)}z^ae^{-z}\,dz \tag 5$$

Putting together $(1)$, $(2)$, and $(5)$ yields the coveted equality

$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty t^ae^{-st}\,dt=\frac{1}{s^{a+1}}\int_0^\infty t^ae^{-t}\,dt}$$

And we are done!

Mark Viola
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  • Math is beautiful! Thank you! – GFauxPas Nov 20 '16 at 00:36
  • Why is the integral along the ray, the second integral on the LHS of (4), tending to zero? Can I use the FToC? – GFauxPas Nov 20 '16 at 14:22
  • You're welcome. My pleasure. And math is beautiful. -Mark – Mark Viola Nov 20 '16 at 18:01
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    The second integral goes to zero because of the exponentially decaying integrand. – Mark Viola Nov 20 '16 at 18:03
  • I graphed the integrand by fixing $a$ and $s$ and in fact the "naive" contour of $\operatorname{Re} + i \operatorname{Im}$ doesn't produce a straight line from the origin to $L\sigma + i L \omega$, it spirals out and then spirals in again. We can use Cauchy-Goursat anyway, because the curve is closed, but it's not simple. – GFauxPas Dec 04 '16 at 00:02
  • Oh, but the spiral intersects the imaginary axis. Is that a problem? – GFauxPas Dec 04 '16 at 17:38
  • @GFauxPas I don't understand your comments. The contour $C$ is comprised of straight line segments by design. There is no spiral. -Mark – Mark Viola Dec 04 '16 at 17:54
  • @Mark the example I'm looking at is $t^{-.5-i}\exp(-(1+i)t)$, with the parameterization $\theta(t) = t$ – GFauxPas Dec 04 '16 at 18:58
  • So, in the development herein, we have $s=1+i$ so that $\sigma =1$ and $\omega =1$. Then, the contour $C$ goes from $0$, to $L$ to $L+iL$ and returns to $0$ (again, we actually have a small arc around the branch point, but this doesn't contribute to the result as the arc length goea to $0$). The integrand is holomorphic in and on $C$. So, what is the issue? – Mark Viola Dec 04 '16 at 19:42
  • I was just looking at the "naive" curve traced out by just parameterising the integrand and choosing an interval $t \in (0..L]$. But since the integrand is analytic, we can choose any contour that begins and ends at the points of interest, is that right? – GFauxPas Dec 04 '16 at 19:59
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    Cauchy's Integral Theorem is valid for any simple rectifiable contour $C$ such that $f$ is analytic in and on $C$. So, we want to choose a contour that is comprised of the straight line path from $0$ and $\sigma L$ and another one that has a straight line path from $0$ to $sL$. We choose then to close with a contour such that it ia easy to show that as L\to \infty$, its contribution goes to $0$. – Mark Viola Dec 04 '16 at 20:15