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We know that the Laplace transform for $f(t)=t^{p}$, $p>0$, is $$ F(s)=\frac{\Gamma(p+1)}{s^{p+1}}. $$ How to derive its inverse Laplace transform by the inversion formula? I mean, if $p=1,2,\dots$ then \begin{align*} f(t)&=\frac{1}{2\pi i}\lim_{T\to\infty}\int_{1-iT}^{1+iT}e^{st}\frac{p!}{s^{p+1}}ds\\ &=\frac{1}{p!}\lim_{z\to0}\frac{d^{p}}{dz^{p}}\left(z^{p+1}e^{zt}\frac{p!}{z^{p+1}}\right)\\ &=t^p, \end{align*} where the second equality follows by using the residue theorem.

However, I have no idea about the case when $p$ is not an integer. Does anyone know how to deal with this case? Thanks.

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Suppose $t>0$. (If $t \leq 0$, we can keep pushing the contour to the right and make the integral as small as we like, so the inverse transform gives $0$ there as it should). We can then change variables immediately to get the $t$-dependence out of the way: if $z=st$, the integral becomes $$ t^p \frac{\Gamma(p+1)}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{e^{z}}{z^{p+1}} \, dz. $$ It remains to evaluate the constant $ \frac{\Gamma(p+1)}{2\pi i} \int_{c-i\infty}^{c+i\infty} \frac{e^{z}}{z^{p+1}} \, dz$. We sweep the contour around to wrap around the branch point at $s=0$ (the integrals over the quarter-circles vanish in the limit by Jordan's lemma). The integral becomes $$ \frac{\Gamma(p+1)}{2\pi i} \int_{\gamma} \frac{e^{z}}{z^{p+1}} \, dz, $$ where $\gamma$ is composed of a line just below the imaginary axis starting at $-\infty$ and ending at $-\varepsilon$, a circle of radius $\varepsilon$ around the branch point at $z=0$, and the line from $-\varepsilon$ back to $-\infty$ just above the imaginary axis. We could try and calculate this directly, but we'll be better off integrating by parts a few times to make the singularity at the branch point better behaved, so we can just take the limits directly. Also, integration by parts is easy with this new contour, because the exponential decays and causes the endpoint terms to vanish. We have $$ \frac{\Gamma(p+1)}{2\pi i} \int_{\gamma} z^{-p-1}e^{z} \, dz = - \frac{\Gamma(p+1)}{2\pi i \,p} \int_{\gamma} -z^{1-p-1}e^{z} \, dz = \frac{\Gamma(p+1-1)}{2\pi i} \int_{\gamma} z^{1-p-1}e^z \, dz \\ = \dotsb = \frac{\Gamma(p+1-n)}{2\pi i} \int_{\gamma} z^{n-p-1} e^{z} \, dz $$

If $n-p>0$, the integrand no longer diverges at $z=0$ (but still has a branch point), so the integral over the circle vanishes as $\varepsilon \to 0$.

Therefore, $$ \int_{\gamma} z^{n-p-1} e^{z} \, dz = \int_{-\infty}^0 z^{n-p-1} e^{z} \, dz + \int_0^{-\infty} z^{n-p-1} e^{z} \, dz $$ Changing variables to $z=xe^{-i\pi}$ in the first and $z=xe^{i\pi}$ in the second gives $$ \int_{-\infty}^0 z^{n-p-1} e^{z} \, dz + \int_0^{-\infty} z^{n-p-1} e^{z} \, dz = -\int_0^{\infty} e^{-i\pi(n-p)} x^{n-p-1} e^{-x} \, dx + \int_0^{\infty} e^{i\pi(n-p)} x^{n-p-1} e^{-x} \, dx \\ = 2i\sin{(n-p)\pi} \int_0^{\infty} x^{n-p-1} e^{-x} \, dx = 2i \Gamma(n-p)\sin{(n-p)\pi}. $$ Thus the whole lot evaluates to $$ t^p \frac{\Gamma(p+1-n)}{2\pi i} 2i \Gamma(n-p)\sin{(n-p)\pi} = \frac{\sin{(n-p)\pi}}{\pi} \Gamma(n-p)\Gamma(p+1-n) $$ But we have the reflection formula $$ \Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{\pi z}}, $$ so the constant $ \frac{\sin{(n-p)\pi}}{\pi} \Gamma(n-p)\Gamma(p+1-n) $ is actually $1$. Therefore the inverse Laplace transform gives $t^p$, as it should.

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  • $\begingroup$ Thanks for your help. I really learn a lot from you. And I have one more question: When $-1<p<0$, is $\frac{\Gamma(p+1)}{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{e^{st}}{s^{p+1}f(s)}ds$ equal to $\frac{1}{\Gamma(-p)}\int_{0}^{\infty}\frac{e^{-st}s^{-p-1}}{f(-s)}ds$ where $f(0)=1$? $\endgroup$ – Tony Apr 17 '17 at 9:17
  • $\begingroup$ You can't tell without more information about $f(s)$. But the integral on the right diverges at zero if $f$ is finite and continuous there (it's $O(s^{-p-1})$). $\endgroup$ – Chappers Apr 17 '17 at 12:02
  • $\begingroup$ Thank you for your reply. Actually, $f(s)=\int_0^{\infty}e^{su}g(u)du<\infty$ for all $s\in\mathbb{R}$ and $f(0)=\int_0^{\infty}g(u)du=1$. I wonder whether the integral can be expressed in terms of something like the right one. $\endgroup$ – Tony Apr 17 '17 at 12:16

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