- #1
myshadow
- 30
- 1
Homework Statement
Find the inverse laplace transform: [tex] {\cal L}^{-1}\{e^{-\sqrt{s-a}c}\} [/tex]
where [itex] a [/itex] and [itex] c [/itex] are constants. [itex] s [/itex] is the complex variable.
Homework Equations
Bromwich integral: [tex] \frac{1}{2\pi{}i}\lim_{T\rightarrow \infty}\int_{\gamma-Ti}^
{\gamma+Ti}e^{st-\sqrt{s-a}c}ds [/tex]
Residue theorem: [tex] \oint f(s) ds=\sum Res(f(s)) [/tex]
The Attempt at a Solution
There is a branch point at [itex] s=a [/itex]. Also, there are no poles. Therefore, the sum of the residues is equal to zero. So I have a keyhole contour centered at [itex] a [/itex] with an outer radius of [itex] R [/itex] and inner radius of [itex] \varepsilon [/itex]. I chose [itex] \gamma=2a [/itex]. I attached a general image of the contour that I used. The contour is centered at [itex] a [/itex].
Therefore, [itex] s=Re^{i\theta}+a [/itex] along the the outer circle,
[itex] s=xe^{\pi i}+a [/itex] going from left to right along the branch cut,
[itex] s=\varepsilon e^{i\theta}+a [/itex] along the the inner circle, and
[itex] s=xe^{-\pi i}+a [/itex] going from right to left along the branch cut.
From Jordan's lemma the integral along the outer radius goes to zero as [itex] R\rightarrow \infty [/itex]. Similarly, the integral along the inner circle (keyhole) goes to zero as [itex] \varepsilon\rightarrow 0 [/itex]
Therefore,
[tex] \frac{1}{2\pi{}i}\lim_{T\rightarrow \infty}\int_{\gamma-Ti}^
{\gamma+Ti}e^{st-\sqrt{s-a}c}ds + \frac{1}{2\pi{}i}\int_{branch cut}=0 [/tex]
[tex] \frac{1}{2\pi{}i}\lim_{T\rightarrow \infty}\int_{\gamma-Ti}^
{\gamma+Ti}e^{st-\sqrt{s-a}c}ds =
\frac{-1}{2\pi{}i}\int_{branch cut}
=\frac{-1}{2\pi{}i}\lim_{\substack{R\rightarrow \infty\\\varepsilon\rightarrow 0}}
(\int_{-R+a}^{-\varepsilon+a}e^{st-\sqrt{s-a}c}ds+
\int_{-\varepsilon+a}^{-R+a}e^{st-\sqrt{s-a}c}ds) [/tex]
[tex] = \frac{-1}{2\pi{}i}(-\int_{a}^{\infty}e^{(a-x)t-\sqrt{x}ic}e^{\pi{}i}dx+\int_{a}^{\infty}e^{(a-x)t+\sqrt{x}ic}e^{-\pi{}i}dx) [/tex]
[tex] = \frac{-1}{2\pi{}i}(\int_{a}^{\infty}e^{(a-x)t-\sqrt{x}ic}dx-\int_{a}^{\infty}e^{(a-x)t+\sqrt{x}ic}dx)
=\frac{e^{ta}}{\pi{}}\int_{a}^{\infty}e^{-xt}\sin{(\sqrt{x}c)}dx [/tex]
And this is where I'm stuck. I think this integral doesn't have a closed form. I can expand the integral into an infinite series but I don't know how to integrate the terms. I tried doing the Inverse Laplace transform in mathematica. The answer it gave me is:
[tex] \frac{ce^{\frac{-c^2}{4t}+at}}{2\sqrt{\pi{}}}t^{\frac{3}{2}} [/tex]
So what I ended up getting doesn't agree with Mathematica. Also I know from tables that the inverse laplace transform of [tex] e^{-\sqrt{s}} [/tex] is [tex] \frac{a}{2\sqrt{\pi{}t^3}}e^{\frac{-a^2}{4t}} [/tex] I really don't know what I did wrong. I don't know what to do from here. Thanks in advance.