Inverse Laplace transform with gamma function? How use the gamma function?

Question

I was trying to solve this inverse Laplace: $\frac{s}{(s+1)^{5/2}}$ using gamma function but somehow I get a bit confused because what I know that gamma function values can be: $$\begin{align}\Gamma(1/2)&=\pi^{1/2}\\ \Gamma(3/2)&=\tfrac{1}{2}\pi^{1/2}\\ \Gamma(5/2)&=\tfrac{3}{4}\pi^{1/2} \end{align}$$ .... But is a correct way to start? $$\begin{split} \quad&=(\mathrm{e}^{-t}) \mathcal{L}^{-1}\{s^{-3/2}\}\\ &=\frac{2\mathrm{e}^{-t}t^{1/2}}{\pi^{1/2}} \end{split}$$

Answer 1

Since

• the inverse Laplace transform of $F(s+a)$ is $\mathrm{e}^{-at}f(t)$ if the inverse Laplace transform of $F(s)$ is $f(t)$,
• the inverse Laplace transform of $s^{-(q+1)}$ is $\tfrac{t^{q}}{\Gamma(q+1)}\theta(t)$, and
• the inverse Laplace transform is linear, we have

$$\begin{split}\mathcal{L}^{-1}\left(\frac{s}{(s+1)^{5/2}}\right)&=\mathcal{L}^{-1}\left(\frac{1}{(s+1)^{3/2}}-\frac{1}{(s+1)^{5/2}}\right)\\ &=\mathcal{L}^{-1}\left(\frac{1}{(s+1)^{3/2}}\right) -\mathcal{L}^{-1}\left(\frac{1}{(s+1)^{5/2}}\right)\\ &=\mathrm{e}^{-t}\left(\mathcal{L}^{-1}(s^{-3/2})-\mathcal{L}^{-1}(s^{-5/2})\right)\\ &=\mathrm{e}^{-t}\left(\tfrac{t^{1/2}}{\Gamma(\tfrac{3}{2})}-\tfrac{t^{3/2}}{\Gamma(\tfrac{5}{2})} \right)\theta(t)\text{.} \end{split}$$

Answer 2

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\begin{align} &\mbox{With}\ c > 0,\ \mbox{note that} \\[2mm] &\int_{c - 1 - \infty\ic}^{c - 1 + \infty\ic}{s \over \pars{s + 1}^{5/2}} \,\expo{ts}{\dd s \over 2\pi\ic} = \expo{-t}\int_{c - \infty\ic}^{c + \infty\ic}\pars{s^{-3/2} - s^{-5/2}} \,\expo{ts}{\dd s \over 2\pi\ic} \label{1}\tag{1} \end{align}

Hereafter, I'll evaluate \begin{align} &\bbox[#ffd,10px]{\ds{% \int_{c - \infty\ic}^{c + \infty\ic}s^{-\nu} \,\expo{ts}{\dd s \over 2\pi\ic}}} = \int_{c - \infty\ic}^{c + \infty\ic}s^{-\nu} \,\expo{ts}{\dd s \over 2\pi\ic} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}\,\,\,& -\int_{-\infty}^{-\epsilon}\pars{-s}^{-\nu}\expo{-\ic\pi\nu}\expo{ts} \,{\dd s \over 2\pi\ic} - \int_{\pi}^{-\pi}\epsilon^{-\nu}\expo{-\ic\nu\theta}{\epsilon\expo{\ic\theta}\ic\dd\theta \over 2\pi\ic} \\[2mm] & - \int_{-\epsilon}^{-\infty}\pars{-s}^{-\nu}\expo{\ic\pi\nu}\expo{ts} \,{\dd s \over 2\pi\ic} \\[5mm] = & -\expo{-\ic\pi\nu}\int_{\epsilon}^{\infty}s^{-\nu}\expo{-ts} \,{\dd s \over 2\pi\ic} - {\sin\pars{\nu\pi} \over \pi\pars{\nu - 1}}\,\epsilon^{1 - \nu} + \expo{\ic\pi\nu}\int_{\epsilon}^{\infty}s^{-\nu}\expo{-ts} \,{\dd s \over 2\pi\ic} \\[5mm] = &\ {\sin\pars{\nu\pi} \over \pi}\bracks{% \int_{\epsilon}^{\infty}s^{-\nu}\expo{-ts}\,\dd s - {\epsilon^{1 - \nu} \over \nu - 1}} = {\sin\pars{\nu\pi} \over \pi}\bracks{% \int_{s\ =\ \epsilon}^{s\ \to\ \infty}\expo{-ts}\, {\dd s^{1 - \nu} \over 1 - \nu} - {\epsilon^{1 - \nu} \over \nu - 1}} \\[5mm] \stackrel{\mrm{IBP}}{=}\,\,\,& -\,{\sin\pars{\nu\pi} \over \pi\pars{\nu - 1}}\,t \int_{\epsilon}^{\infty}s^{1 - \nu}\expo{-ts}\,\dd s \,\,\,\stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\large\to}\,\,\, -\,{\sin\pars{\nu\pi} \over \pi\pars{\nu - 1}}\,t^{\nu - 1}\, \Gamma\pars{2 - \nu}\label{2}\tag{2} \end{align}

whenever $\ds{\Re\pars{\nu} < 2}$. The integral, which involves the factor $\ds{s^{-5/2}}$, $\color{red}{\texttt{diverges}}$ !!!.

Nothe that $\ds{\nu = {3 \over 2}}$ ( with expression \eqref{2} ) yields the result $\ds{2\root{t \over \pi}}$.