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Does anyone know how to calculate the Inverse Laplace transform of $\;\;\dfrac{(s+1)e^{-\pi s}}{s^2 + s + 1}\;\,$ ?

I've tried it and got (u is the unit step function):


But this looks wrong somehow. Please can you clarify whether I'm correct and, if not, perhaps guide me in the right direction. I've spent a long, long time on this problem!

Thank you in advance and Happy New Year!

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is this the function? $$ \frac{(s+1)e^{-\pi s}}{s^2 + s + 1}$$ –  Santosh Linkha Jan 11 '13 at 1:41
yes, that is. Do you know what the inverse is? Am I correct above? –  user56866 Jan 11 '13 at 1:45
Thank you experimentX for re-formatting my question –  user56866 Jan 11 '13 at 1:46
you can't have $s$ terms after taking inverse transform –  Santosh Linkha Jan 11 '13 at 1:48
but if you separate e^pi(s) and leave the fraction as (s+1)/((s+1)^2-S) surely this is the standard transform for cos(st)? –  user56866 Jan 11 '13 at 1:53

2 Answers 2

First do a completion of squares on the denominator $s^2+s+1=(s+1/2)^2+(\sqrt 3 /2)^2$. Then break up the numerator as a linear combination of the two bases on the denominator $s+1=(s+1/2)+1/\sqrt 3 (\sqrt3/2)$. Now you have

${{s+1}\over{s^2+s+1}}= {{{s+1/2}\over{(s+1/2)^2+(\sqrt 3 /2)^2}}+{{1/\sqrt3}{{\sqrt3/2}\over{(s+1/2)^2+(\sqrt 3 /2)^2}}}}$.

Now you look up each of above fractions in your table to get

$e^{-t/2 }\cos(\sqrt3 t/2)+{1/\sqrt3}e^{-t/2} \sin(\sqrt3 t/2)$.

Now you bring in $e^{-\pi s}$. It gives $U_\pi (t)$ and a shift of $\pi$ in $t$ to produce

$U_\pi (t) e^{-(t-\pi)/2} \left[ \cos(\sqrt3 (t-\pi)/2)+{1/\sqrt3} \sin(\sqrt3 (t-\pi)/2)\right]$.

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Thank you so much for explaining this so clearly. –  user56866 Jan 11 '13 at 11:06

Forgive me for doing this without a picture of the contour for now. I can add later if you wish.

The inverse Laplace transform we seek is

$$ \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: e^{s t} \frac{s+1}{s^2+s+1} e^{-\pi s}$$

$$ \frac{1}{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \: \frac{s+1}{s^2+s+1} e^{(t-\pi) s}$$

We consider first the case $t>\pi$. In this case, we use a contour from the $\Re{s} = c$, where $c>0$, and the portion of the circle $|s|=R$ that contains the poles of the integrand. These poles are at $s=\frac{1}{2} \pm i \frac{\sqrt{3}}{2}$.

We use the Residue Theorem, which states that the integral around the closed contour described above is equal to $i 2 \pi$ times the sum of the residues of the poles contained within the contour. I can go into more detail here if you want, but the sum of the residues at the two poles above is

$$ e^{-\frac{1}{2} (t-\pi)} \left [ \cos{ \left [ \frac{\sqrt{3}}{2} (t-\pi) \right ] } + \frac{1}{\sqrt{3}} \sin{ \left [ \frac{\sqrt{3}}{2} (t-\pi) \right ] } \right ] $$

For $t<\pi$, we must use a contour in which the circular portion goes to the right of the line $\Re{s} = c$. As there are no poles within this contour, the integral is zero here.

Therefore, the inverse Laplace transform is given by

$$ e^{-\frac{1}{2} (t-\pi)} \left [ \cos{ \left [ \frac{\sqrt{3}}{2} (t-\pi) \right ] } + \frac{1}{\sqrt{3}} \sin{ \left [ \frac{\sqrt{3}}{2} (t-\pi) \right ] } \right ] U(t-\pi) $$

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Factor of 1/2 was not correct, removed. –  Ron Gordon Jan 11 '13 at 2:27
I don't need the contour, thank you for your detailed answer, I really appreciate it –  user56866 Jan 11 '13 at 11:06

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