# Inverse Laplace transform of $s^{k}$

How can I find the inverse Laplace transform of $s^{k}$ where $k$ is non-integer and negative?

I know that $$\mathcal{L}^{-1}[s^k] = \frac{1}{2\pi i}\int e^{st} s^k ds$$

and since we have singularity at $0$ and $k$ is non-integer, I suppose I have to take a branch cut along the negative real axis, i.e use the principle log $e^{Log(s^{k})}$. Then we integrate along the contour $\Gamma$:

$$\oint_{\Gamma} e^{st} s^k ds = 0$$

where $\Gamma$ is: (Image credit to @ron-gordon: https://math.stackexchange.com/a/504788/98393)

Then integral on $C_4, C_2, C_6$ vanish, leaving us with

$$0 = \int_{C_3 + C_5 + C_1} e^{st} s^k ds \Rightarrow 2\pi i \mathcal{L}^{-1}[s^k] = \int_{C_1}e^{st} s^k ds = - \int_{C_3 + C_5} e^{st} s^k ds$$

This is where I get stuck, I'm not sure how to compute the integrals... and I'm not even sure if I am on the right track. Could someone please give me some hints?

Note: this is a practice question not a homework question.

One option, in particularly if one knows (roughly) what to expect is to attack it from the other side. If we didn't already know what to expect, we could shape our expectation by considering the case where $k = -m$ is a negative integer. Then the plain semicircular contour works, and we obtain

$$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} e^{st}s^{-m}\,ds = \operatorname{Res}\left(e^{zt}z^{-m};0\right) = \frac{t^{m-1}}{(m-1)!},$$

which suggests we look at $\frac{t^{m-1}}{\Gamma(m)}$ also for non-integer $m > 0$. Indeed, considering real $s > 0$ in the Laplace integral to have no problems with substitution, we find

$$\int_0^\infty t^{m-1}e^{-st}\,dt = \frac{1}{s^m}\int_0^\infty u^{m-1}e^{-u}\,du = \frac{\Gamma(m)}{s^m},$$

and by the identity theorem that holds for all $s$ with $\operatorname{Re} s > 0$, so indeed $\frac{t^{m-1}}{\Gamma(m)} = \frac{t^{-(k+1)}}{\Gamma(-k)}$ is the inverse Laplace transform of $s^k$ for any $k < 0$, integer or not.

When we use the contour wrapping the branch-cut, note that the integral over the part $C_4$ only tends to $0$ in the limit if $-1 < k < 0$.

For the integral over $C_3$, in the limit moving the path to the negative real axis, we get $\operatorname{Log} s = \log \lvert s\rvert + i\pi$, so

\begin{align} \int_{C_3} e^{st}s^k\,ds &\to \int_{-R}^{-\varepsilon} e^{st} \exp\left(k(\log (-s)+i\pi)\right)\,ds\\ &= e^{ik\pi}\int_{-R}^{-\varepsilon} e^{st}\lvert s\rvert^k\,ds. \end{align}

Similarly, on $C_5$ the logarithm becomes $\log \lvert s\rvert -i\pi$, and

\begin{align} \int_{C_5}e^{st}s^k\,ds &\to -\int_{-R}^{-\varepsilon} e^{st}\exp\left(k(\log(-s)-i\pi)\right)\,ds\\ &= -e^{-ik\pi}\int_{-R}^{-\varepsilon} e^{st}\lvert s\rvert^k\,ds. \end{align}

Together,

$$\int_{C_3+C_5} e^{st}s^k\,ds \to 2i\sin(k\pi)\int_{\varepsilon}^R e^{-xt}x^k\,dx.$$

For $-1 < k < 0$, we can take the limit as $\varepsilon\to 0$ and $R\to\infty$ and obtain

$$\frac{1}{2\pi i} \int_{C_3+C_5} e^{st}s^k\,ds \to \frac{\sin (k\pi)}{\pi}\frac{\Gamma(k+1)}{t^{k+1}}.$$

By Euler's reflection formula for the $\Gamma$-function and the functional equation $\Gamma(z+1) = z\Gamma(z)$,

$$\frac{\sin (k\pi)}{\pi} = \frac{1}{\Gamma(k)\Gamma(1-k)} = \frac{1}{\Gamma(k)(-k)\Gamma(-k)} = -\frac{1}{\Gamma(k+1)\Gamma(-k)},$$

so

$$\frac{1}{2\pi i} \int_{C_3+C_5} e^{st}s^k\,ds \to -\frac{t^{-(k+1)}}{\Gamma(-k)},$$

and since the integral over the entire contour vanishes by Cauchy's integral theorem,

$$\frac{1}{2\pi i}\int_{c-i\infty}^{c+i\infty} e^{st}s^k\,ds = \frac{t^{-(k+1)}}{\Gamma(-k)},$$

For $k$ a negative integer, the integrals over $C_3$ and $C_5$ cancel in the limit, and the integral over $C_4$ becomes the residue in $0$ (times $2\pi i$). For $k < -1$ not an integer, the limit as $\varepsilon\to 0$ of the integral over $C_4$ does not exist, nor does the limit of the integral over $C_3+C_5$. But for the integral over $C_3+C_4+C_5$, by $\lfloor -k\rfloor$ integrations by part, we obtain the result.
• Thank you for your detailed answer! I have a question: for $k$ a negative integer, can we still take the branch cut? I remember that we only need to take the branch cut if there's branch points. So if we don't have a branch cut, how can we form a closed contour?
• If $k$ is a negative integer, the same contour works (but, the standard semicircle closing the contour would in that case work too). For integral $k$, since there is no actual branch cut, the values on $C_3$ and $C_5$ are the same when these are moved to the negative real axis, and the integrals cancel. We are then left with the closed contour consisting of the line segment parallel to the imaginary axis and the large semicircle, and a small circle around the singularity, connected by a line traversed once in each direction. The end result is that the integral over the straight line equals Commented Apr 20, 2014 at 21:46
• $2\pi i$ times the residue in $0$, since the integral over the semicircle vanishes in the limit, the integrals over the connecting line cancel, and the integral over the circle is $-2\pi i$ times the residue. Commented Apr 20, 2014 at 21:48