Inverse Laplace transform of a function having algebraic singularity

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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