Inverse Laplace transform of a function having algebraic singularity

Submitted by Structure on Sat, 12/22/2007 - 22:59.

Laplace inverse transform of $F(s)=e^{-\sqrt s}$
This function has two singular points, zero and $\infty$ none of them isolated.
We use the differential equation satisfied by F:

$4sF"(s)+2F'(s)-F(s)=0$

Let $F(s)=\mathcal{L}(f(t))(s)$
Then
$F'(s)=- \mathcal{L}(tf(t))(s)$
$F"(s)= \mathcal{L}(t^2f(t))(s)$
$sF"(s)= \mathcal{L}(\frac{d (t^2f(t))}{dt})(s)$
So

$4sF"(s)+2F'(s)-F(s)=\mathcal{L}(4\frac{d (t^2f(t))}{dt}-2tf(t)-f(t))(s)=0$

Or

$4\frac{d (t^2f(t))}{dt}-2tf(t)-f(t))=0$

Or

$4t^2f'(t)+6tf(t)-f(t)=0$

Or

$\frac{f'(t)}{f(t)}=\frac{1}{4t^2}-\frac{3}{2t}$

$\ln f=-\frac{1}{4t}-\frac{3}{2}\ln t+\ln c$

or

$f(t)=c \:t^{-\frac{3}{2}}e^{-\frac{1}{4t}$

Now

$tf(t)=c \:t^{-\frac{1}{2}}e^{-\frac{1}{4t}}h(t)$

and

$\mathcal{L}(tf(t)(s)=-F'(s)=\frac{1}{2\sqrt s}F(s)=\frac{1}{2\sqrt s}e^{-\sqrt s}$

But $tf(t)$ is equivalent at $\+\infty$ to $g(t)=c \:t^{-\frac{1}{2}}h(t)$ and then their Laplace transform are equivalent at s=0 But

$G(s)=\mathcal{L}(g(t))(s)=c\frac{\sqrt \pi}{\sqrt s}$

From

$c\frac{\sqrt \pi}{\sqrt s}\: equivalent \: \frac{1}{2\sqrt s}e^{-\sqrt s}$

we have

$c=\frac{1}{2\sqrt{\pi}}$

.
Finally we have

$f(t)=\frac{1}{2\sqrt{\pi}}\:t^{-\frac{3}{2}}e^{-\frac{1}{4t}}h(t)$

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