Calculating Inverse Laplace Transform of ((1+s/wh)/(1+s/wb))^n

Question

I try to compute the inverse Laplace Transform of $\left( \frac{1+s/\omega_{h}}{1+s/\omega_{b}}\right) ^{n}$ with $0<\omega_{b}<\omega_{h}$ and $0<n<1$. As this function has 2 branch points, one at $-\omega_{b}$ and the other at $-\omega_{h}$, I considered the following integration path:considered path to apply residue theorem.

I found that all the residue are null (perhaps wrong) and there is a contribution only of path $\gamma_{3}$ and $\gamma_{4}$. Thus the result obtained is $\frac{sin(n\pi)}{\pi}\int_{\omega_{b}}^{\omega_{h}} \left( \frac{1-u/\omega_{h}}{u/\omega_{b}-1}\right) ^{n}e^{-tu} du$ but this is wrong, something is missing in this result, somethink like a dirac function multiplied by $\frac{\omega_{b}}{\omega_{h}}$ that I do not know how to make appear. Thank you in advance to those who could help me.

Answer 1

First, pull a factor of $(\omega_b/\omega_h)^n$ outside and consider the integral

$$\oint_C dz \, (z+\omega_h)^n (z+\omega_b)^{-n} e^{z t} $$

where $C$ is the modified Bromwich integral that detours about the negative axis around the branch points at $z=-\omega_b$ and $z=-\omega_h$ when $t \gt 0$. We may then neglect the integrals about the large circular arcs of radius $R$ which vanish as $R \to \infty$ and the small circular arcs about the branch points of radius $\epsilon$ as $\epsilon \to 0$. The contour integral is then

$$\int_{c-i \infty}^{c+i \infty} ds \, (s+\omega_h)^n (s+\omega_b)^{-n} e^{s t} + e^{i \pi} \int_{\infty}^{\omega_b} dx \, (x-\omega_h)^n (x-\omega_b)^{-n} e^{-x t} \\ + e^{i \pi} \int_{\omega_h}^{\omega_b} (\omega_h-x)^n e^{-i \pi n} (x-\omega_b)^{-n} e^{-x t} + e^{-i \pi} \int_{\omega_b}^{\omega_h} (\omega_h-x)^n e^{i \pi n} (x-\omega_b)^{-n} e^{-x t} \\ + e^{-i \pi} \int_{\omega_b}^{\infty} dx \, (x-\omega_h)^n (x-\omega_b)^{-n} e^{-x t} $$

where $c \gt -\omega_b$. Note that the factor of $-1$ was replaced with $e^{i \pi}$ above the branch cut $\operatorname{Re}{z} < -\omega_b$ and $e^{-i \pi}$ below the branch cut. The phase factors in the second and fifth integrals vanish, so that the second and fifth integrals cancel.

By Cauchy's theorem, the contour integral is zero. We may then write down the inverse Laplace transform in terms of a real integral as follows:

$$\begin{eqnarray} \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, (s+\omega_h)^n (s+\omega_b)^{-n} e^{s t} &= \frac{\sin{\pi n}}{\pi} \int_{\omega_b}^{\omega_h} (\omega_h-x)^n (x-\omega_b)^{-n} e^{-x t} \\ &= \frac{\sin{\pi n}}{\pi} e^{-\omega_b t} \int_0^{\omega_h-\omega_b} dx \, x^n (\omega_h-\omega_b-x)^{-n} e^{-x t} \\ &= \frac{\sin{\pi n}}{\pi} e^{-\omega_b t} (\omega_h-\omega_b) \int_0^1 du \, u^n (1-u)^{-n} e^{-(\omega_h-\omega_b) t u} \end{eqnarray} $$

The latter integral is a confluent hypergeometric function:

$$\int_0^1 du \, u^n (1-u)^{-n} e^{- \beta u} = \frac{\pi n}{\sin{\pi n}} {}_1F_1(1+n,2,-\beta) $$

Thus, the ILT of the specified function is, for $t \gt 0$,

$$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \left (\frac{1+s/\omega_h}{1+s/\omega_b} \right )^n e^{s t} = n (\omega_h-\omega_b) \left (\frac{\omega_b}{\omega_h} \right )^n e^{-\omega_b t} {}_1F_1(1+n,2,-(\omega_h-\omega_b) t)$$

For $t \lt 0$, it should be clear that the contour $C$ is closed to the right. Because the contour encloses no singularities, the ILT is zero in this case.

Answer 2

Dea Jean-Marie, Dear Ron,

First of all, thank you very much for your kind help.

Ron, please note that I considered $\left(\frac{ωh+s}{ωb+s}\right)^{n}$ and not $\frac{(ωh+s)^{n}}{(ωb+s)^{n}}$ as in your first integral. Could you please confirm that the two expressions are equivalent as you consider a path that exclude the segment [-$\infty$, -wb].

Also, I do not understand why you consider a branch cut that exclude the real axis from $-\infty$ to -wb? The function I considered (the ratio raised to the power n and not the ration of two terms each raised to the power n) is defined on [$-\infty$, -wh]. Why the path I proposed in my question is not correct?

Answer 3

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With $\ds{\quad 0 < \omega_{b} < \omega_{h}\quad}$ and $\ds{\quad 0 < n < 1\quad}$.

\begin{align} &\int_{\pars{-\omega_{b}}^{+} - \infty\ic}^{\pars{-\omega_{b}}^{+} + \infty\ic} \pars{1 + s/\omega_{h} \over 1 + s/\omega_{b}}^{n}\expo{ts} {\dd s \over 2\pi\ic} = \pars{\omega_{b} \over \omega_{h}}^{n}\int_{\pars{-\omega_{b}}^{+} - \infty\ic}^{\pars{-\omega_{b}}^{+} + \infty\ic} \pars{s + \omega_{h} \over s + \omega_{b}}^{n}\expo{ts}{\dd s \over 2\pi\ic} \\[5mm] = & \pars{\omega_{b} \over \omega_{h}}^{n}\expo{-\omega_{b}t} \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} \pars{s - \omega_{b} + \omega_{h} \over s}^{n}\expo{ts}{\dd s \over 2\pi\ic} \\[5mm] = &\ \pars{\omega_{b} \over \omega_{h}}^{n}\nu \expo{-\omega_{b}t}\bbox[#ffd,10px]{\ds{% \int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} \pars{s + 1}^{n}\,s^{-n}\expo{\nu ts}{\dd s \over 2\pi\ic}}}\,,\qquad \nu \equiv \omega_{h} - \omega_{b} \end{align}

---

\begin{align} &\bbox[#ffd,10px]{\ds{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} \pars{s + 1}^{n}\,s^{-n}\expo{\nu ts}{\dd s \over 2\pi\ic}}} \\[5mm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\sim}&\ -\int_{-\infty}^{-1} \pars{-s - 1}^{n}\expo{n\pi\ic}\pars{-s}^{-n}\expo{-n\pi\ic}\expo{\nu ts} \,{\dd s \over 2\pi\ic} - \int_{-1}^{-\epsilon}\pars{s + 1}^{n}\pars{-s}^{-n}\expo{-n\pi\ic}\expo{\nu ts} \,{\dd s \over 2\pi\ic} \\[3mm] & - \int_{\pi}^{-\pi}\epsilon^{-n}\expo{-\ic n\theta} \,{\epsilon\expo{\ic\theta}\ic\,\dd\theta \over 2\pi\ic} \\[3mm] & - \int_{-\epsilon}^{-1}\pars{s + 1}^{n}\pars{-s}^{-n}\expo{n\pi\ic}\expo{\nu ts} \,{\dd s \over 2\pi\ic} - \int_{-1}^{-\infty} \pars{-s - 1}^{n}\expo{-n\pi\ic}\pars{-s}^{-n}\expo{n\pi\ic}\expo{\nu ts} \,{\dd s \over 2\pi\ic} \\[1cm] = &\ -\expo{-n\pi\ic}\int_{\epsilon}^{1}\pars{1 - s}^{n}s^{-n}\expo{-\nu ts} \,{\dd s \over 2\pi\ic} + {\sin\pars{n\pi} \over \pars{n - 1}\pi}\,\epsilon^{1 - n} + \expo{n\pi\ic}\int_{\epsilon}^{1}\pars{1 - s}^{n}s^{-n}\expo{-\nu ts} \,{\dd s \over 2\pi\ic} \\[5mm] = &\ {\sin\pars{n\pi} \over \pi} \int_{\epsilon}^{1}\pars{1 - s}^{n}s^{-n}\expo{-\nu ts}\,\dd s - {\sin\pars{n\pi} \over \pars{n - 1}\pi}\,\epsilon^{1 - n} \\[1cm] \stackrel{\mbox{IBP}}{=}\,\,\, &\ {\sin\pars{n\pi} \over \pi}\bracks{% -\,{\epsilon^{-n + 1}\pars{1 - \epsilon}^{n}\expo{-\nu t\epsilon} \over -n + 1} -\int_{\epsilon}^{1}{s^{-n + 1} \over -n + 1} \braces{\partiald{}{s}\bracks{\pars{1 - s}^{n}\expo{-\nu ts}}}\,\dd s} \\[3mm] &\ - {\sin\pars{n\pi} \over \pars{n - 1}\pi}\,\epsilon^{1 - n} \\[1cm] \stackrel{\mrm{as}\ \epsilon\ \to\ 0^{+}}{\to}\,\,\,& {\sin\pars{n\pi} \over \pars{n - 1}\pi}\int_{0}^{1}s^{1 - n}\bracks{% n\pars{1 - s}^{n - 1}\pars{-1}\expo{-\nu ts} + \pars{1 - s}^{n}\expo{-\nu ts}\pars{-\nu t}}\dd s \\[1cm] = &\ -\,{n\sin\pars{n\pi} \over \pars{n - 1}\pi} \int_{0}^{1}\expo{-\nu ts}s^{1 - n}\pars{1 - s}^{n - 1}\,\dd s - {\sin\pars{n\pi} \over \pars{n - 1}\pi}\,\nu t \int_{0}^{1}\expo{-\nu ts}s^{1 - n}\pars{1 - s}^{n}\,\dd s \end{align}

Note that $\ds{\pars{~\tilde{\mrm{F}}_{1}\ \mbox{is the}\ Regularized\ Hypergeometric\ {}_{1}\mrm{F}_{1}~}}$:

$$ \left\{\begin{array}{rcl} \ds{\int_{0}^{1}\expo{-\nu ts}s^{1 - n}\pars{1 - s}^{n - 1}\,\dd s} & \ds{=} & \ds{\Gamma\pars{n}\Gamma\pars{2 - n}\,\tilde{\mrm{F}}_{1}\pars{2 - n;2;-\nu t}} \\ & \ds{=} & \ds{-\,{\pars{n - 1}\pi \over \sin\pars{\pi n}} \,\tilde{\mrm{F}}_{1}\pars{2 - n;2;-\nu t}} \\[5mm] \ds{\int_{0}^{1}\expo{-\nu ts}s^{1 - n}\pars{1 - s}^{n}\,\dd s} & \ds{=} & \ds{\Gamma\pars{1 + n}\Gamma\pars{2 - n} \,\tilde{\mrm{F}}_{1}\pars{2 - n;3;-\nu t}} \\ & \ds{=} & \ds{-\,{n\pars{n - 1}\pi \over \sin\pars{\pi n}} \,\tilde{\mrm{F}}_{1}\pars{2 - n;3;-\nu t}} \end{array}\right. $$ such that \begin{align} &\bbox[#ffd,10px]{\ds{\int_{0^{+} - \infty\ic}^{0^{+} + \infty\ic} \pars{s + 1}^{n}\,s^{-n}\expo{\nu ts}{\dd s \over 2\pi\ic}}} = n\,\tilde{\mrm{F}}_{1}\pars{2 - n;2;-\nu t} + n\nu t\,\tilde{\mrm{F}}_{1}\pars{2 - n;3;-\nu t} \end{align}

---

\begin{align} &\int_{\pars{-\omega_{b}}^{+} - \infty\ic}^{\pars{-\omega_{b}}^{+} + \infty\ic} \pars{1 + s/\omega_{h} \over 1 + s/\omega_{b}}^{n}\expo{ts} {\dd s \over 2\pi\ic} \\[5mm] = &\ \bbx{\ds{% n\pars{\omega_{b} \over \omega_{h}}^{n}\pars{\omega_{h} - \omega_{b}} \expo{-\omega_{b}t}\bracks{% \tilde{\mrm{F}}_{1}\pars{2 - n;2;-\nu t} + \pars{\omega_{h} - \omega_{b}}t \,\tilde{\mrm{F}}_{1}\pars{2 - n;3;-\pars{\omega_{h} - \omega_{b}}t}}}} \end{align}