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Find the inverse Laplace transform

$F(t)=\mathcal{L}^{-1}(s^{-\frac{1}{2}}e^{-\frac{1}{s}})$

using each of the following techniques:

  1. Expand the exponential in a Taylor series about s=∞, and take inverse Laplace transforms term by term (this is allowable since the series is uniformly convergent.).

  2. Sum the resultant series in terms of elementary functions.

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  • $\begingroup$ That looks doable. What part have you problems with? $\endgroup$ Commented Oct 20, 2013 at 18:17
  • $\begingroup$ Expanding the exponential in a Taylor series about s=∞. $\endgroup$
    – Ravi
    Commented Oct 20, 2013 at 18:19
  • $\begingroup$ Ah. You know $e^z = \sum\limits_{n=0}^\infty \frac{1}{n!}z^n$, I suppose. Just insert $z = -\frac1s$ and you have it. $\endgroup$ Commented Oct 20, 2013 at 18:21
  • $\begingroup$ I am just stuck here: after expanding I am getting the series as :1/s^.5 -1/s^1.5 + 1/s^2.5 ..so on.Now, I think I just need to manipulate it, so that I can use this formula:1/s^(n+1)=(t^n)/n!..But somehow could not able to translate into that form.Any help please? $\endgroup$
    – Ravi
    Commented Oct 20, 2013 at 19:36
  • $\begingroup$ What is the inverse Laplace transform of $s^{-\alpha}$ generally? $\endgroup$ Commented Oct 20, 2013 at 19:39

3 Answers 3

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We know that $$ \displaystyle \mathcal{L} \{t^{n-1}\} = \frac{\Gamma(n)}{s^{n}}, n>0 ,s>0 $$ $$ \frac{1}{s^n} = \frac{\displaystyle \mathcal{L} \{t^{n-1}\}}{\Gamma(n)} $$ $$ \displaystyle \mathcal{L^{-1}} \{\frac{1}{s^n}\} = \frac{t^{n-1}}{\Gamma(n)} $$ Therefore $$ \displaystyle \mathcal{L^{-1}} \{ s^{-1/2} e^{-1/s} \} = \displaystyle \mathcal{L^{-1}} \{ \frac{1}{s^{1/2}} - \frac{1}{s^{3/2}} + \frac{1}{(2!s^{5/2})} + ... + (-1)^{n}\frac{1}{n! s^{n+\frac{1}{2}}} \} $$ $$ = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \displaystyle \mathcal{L^{-1}} \{\frac{1}{s^{n+\frac{1}{2}}}\} $$ $$ = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!} \frac{t^{n-\frac{1}{2}}}{\Gamma(n+\frac{1}{2})} $$ But after this I don't know how to simplify

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you should use the gamma function to find the inverse of each of the 1/sqrt(s) instead of the basic integer factorial. then take out a common factor of sqrt(t) and find a general series sum from n=0 to infinity. you also seem to be missing a 1/n! in the series.

then use the given gamma formula from the standard formulae to sub for the Gamma(n+1/2) function. you can then use wolfram to get a standard function that the series will converge to.

Then suggest what I can do for part b :)

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  • $\begingroup$ I have done this:We know, Taylor Series:〖 e〗^z=∑_(n=0)^∞▒1/n! z^n Here, we take z=-1/s So, L^(-1) {s^(-1/2) e^(-1/s)} =L^(-1) {〖(-z)〗^(-1/2) 1/n! z^n} =- L^(-1) {1/n! 1/√z z^n} =-[L^(-1) { z^(-1/2) z^1} - L^(-1) {1/2! z^(-1/2) z^2} + … ] =-[L^(-1) {1/1! z^(1/2) } - L^(-1) {1/2! z^(3/2) } + L^(-1) {1/2! z^(5/2) }-…] $\endgroup$
    – Ravi
    Commented Oct 21, 2013 at 11:21
  • $\begingroup$ Now what?I am not sure about Gama Function?Is there anyway I can do inverse Laplace transform of s−α generally? $\endgroup$
    – Ravi
    Commented Oct 21, 2013 at 11:22
  • $\begingroup$ You should have $s^{-1/2}*exp(-1/s) = 1/s^{1/2} - 1/s^{3/2} + 1/(2!s^{5/2})+...$ etc.. then $L^{-1} {1/s^n} = t^{n-1}/Gamma(n)$ from Spiegel. Apply the inverse to the series and solve one by one. then take out a common factor of $1/t^{0.5}$ $\endgroup$
    – Boris
    Commented Oct 21, 2013 at 13:31
  • $\begingroup$ The series sum should be $\sum_{n=0}^\infty (-t)^n/(Gamma(n+1/2)*n!)$ Enter the given formula for gamma(n+1/2), simplify a little and enter that series into wolfram. you should get a fairly standard function $\endgroup$
    – Boris
    Commented Oct 21, 2013 at 13:39
  • $\begingroup$ How you are deducing that the series sum should be ∑∞n=0(−t)n/(Gamma(n+1/2)∗n!)?Also Gamma (n+1/2),I could not find any formula for that.But as in Wolfram I got:(2 (n+0.5)!)/(2 n+1) $\endgroup$
    – Ravi
    Commented Oct 21, 2013 at 16:43
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$\frac{\cos 2(\sqrt{t})}{\sqrt{\pi t}}$

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  • $\begingroup$ I know the answer can be determined directly from the Wolfram, but that is not what I am suppose to do.:) $\endgroup$
    – Ravi
    Commented Oct 22, 2013 at 8:54
  • $\begingroup$ It is also in the tables from Spiegel $\endgroup$
    – anon
    Commented Oct 22, 2013 at 22:49
  • $\begingroup$ I know,but I have to 'proof' it! $\endgroup$
    – Ravi
    Commented Oct 23, 2013 at 18:45

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