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Prove the following Laplace transforms:

(a) $ \displaystyle{\mathcal{L} \{ t^{-1/2} \} = \sqrt{\frac{ \pi}{s}}} ,s>0 $

(b) $ \displaystyle{\mathcal{L} \{ t^{1/2} \} =\frac{1}{2s} \sqrt{\frac{ \pi}{s}}} ,s>0 $

I did (a) as following:

(a) $ \displaystyle{\mathcal{L} \{ t^{-1/2} \} = \int_{0}^{\infty} e^{-st} t^{-1/2}dt }$. Substituting $st=u$ and using the fact that $\displaystyle { \int_{0}^{\infty} e^{-u^2}du=\sqrt{\pi} }$ we are done.

Is there a similar way about (b)? Can we make a substitution to get in (a)?

edit: I know the formula $ \displaystyle \mathcal{L} \{ t^n \} = \frac{\Gamma (n+1)}{s^{n+1}}, n>-1 ,s>0$ , but I would like to see a solution without this.

Thank's in advance!

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up vote 3 down vote accepted

Like I mentioned earlier, there is the rule $\mathcal{L}\{tf(t)\}=-F'(s)$, here applicable with $f(t)=t^{-1/2}$.

Or just directly apply $d/ds$ to part (a). Integration by-parts is equivalent ($u=t^{1/2},dv=e^{-ts}dt$).

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Thank's for you help once again! I think I didn't get used this rule. I have to pracrice more on this. – passenger Mar 26 '12 at 23:59

For $t^{-1/2}$ we have

$$F(s)=\int\limits_0^\infty e^{-st} t^{-1/2}dt$$ Now make $st = u$ so that

$$F(s)=s^{-1/2} \int\limits_0^\infty e^{-u} u^{-1/2}du$$

Since the integral is $\Gamma(1/2)$ we get

$$F(s)=s^{-1/2} \sqrt \pi=\sqrt{\frac{\pi}{s}}$$

Why don't you want to prove the general case? Use the best tools you have when you can. We have

$$\mathcal{L}(t^n)=\int\limits_0^\infty e^{-st}t^n dt$$

We make $st = u$ and get

$$\mathcal{L}(t^n)=\frac{1}{s^{n+1}}\int\limits_0^\infty e^{-u}u^n du$$



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What happened to if $u = st$ then $du = sdt$. So $dt = \frac{du}{s}$ are we missing an s in the denominator? Can someone explain this? – Daniel Miladinovich May 5 '14 at 1:59

simplification of the above

st = u

s dt = du

dt = du/s

1/s integral(0- infinity ) e^-u (u/s)^-1/2 du

s^1/2 / s^1 = 1/s^1/2 therefore the answer is root pi by s since the multiplying term is s^1/2 again.

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Here's a MathJax tutorial :) – Shaun Jul 9 '14 at 9:34

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