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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!

Pedro
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passenger
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4 Answers4

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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

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

Pedro
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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? – Dr. Dan May 05 '14 at 01:59
  • @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. – snulty Oct 10 '16 at 15:51
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 ($u=t^{1/2},dv=e^{-ts}dt$).

Community
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anon
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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
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f (t)=t^-1/2 ANSWER: clearly,f is not defined at t=0,but it will be shown that laplace of (t^1/2) EXITS. By definition we have Let st =x then s dt=dx or. dt=1/s dx

t^-1/2=(x/s)^-1/2 substitute these values to formula of laplace 1/(s)^1/2 (sign of integral)0 to infinity e^-x.x^-1/2 dx 1/(s)1/2(1/2) =(pie/s) ANSEWR

freeha jamshaid
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-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.

sapre
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