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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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    @PedroTamaroff it's okay. I figured out that you're right, but it involves some additional substitutions,2016-12-31

4 Answers 4

-1

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.

  • 1
    [Here's a MathJax tutorial](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) :)2014-07-09
9

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

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

Thus

$\mathcal{L}(t^n)=\frac{\Gamma(n+1)}{s^{n+1}}$

  • 0
    @DanielMiladinovich You probably either understand it by now, or have forgotten but there's also the $t^{-1/2}=u^{-1/2}s^{1/2}$, and that is cancelled by the $1/s$ to give $s^{-1/2}$ outside the integral.2016-10-10
5

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.2012-03-26