# Inverse Laplace Transform

• May 22nd 2011, 01:54 PM
MathMan170
Inverse Laplace Transform
Sorry if this is in the wrong section.

Basically I am trying to find the inverse laplace transform of

$F(s)=\frac{s}{\left ( s+1 \right )^2+1}$

using the bromiwich integral and the standard bromwich contour. I therefore used the residue theorem to evaluate

$\frac{1}{2\pi i}\int _\gamma \frac{e^{zt} z}{\left ( z+1 \right )^2+1} dz$

and I found the answer to be

$\left ( \frac{1}{2}+\frac{i}{2} \right )e^{\left ( -1+i \right )t}+\left ( \frac{1}{2}-\frac{i}{2} \right )e^{\left ( -1-i \right )t}$

Which I believe is right, however when I try to show that the curved part of the contour goes to zero as R tends to infinity I find that it does not, and so my answer if not what I want.

Now I have been looking at ways around this and jordans lemma keeps appearing, although I really have no idea how to apply it (or even if this is the right approach), could someone please explain it or point me in the right direction?

Thanks,

Lewis.
• May 22nd 2011, 07:22 PM
TheEmptySet
Quote:

Originally Posted by MathMan170
Sorry if this is in the wrong section.

Basically I am trying to find the inverse laplace transform of

$F(s)=\frac{s}{\left ( s+1 \right )^2+1}$

using the bromiwich integral and the standard bromwich contour. I therefore used the residue theorem to evaluate

$\frac{1}{2\pi i}\int _\gamma \frac{e^{zt} z}{\left ( z+1 \right )^2+1} dz$

and I found the answer to be

$\left ( \frac{1}{2}+\frac{i}{2} \right )e^{\left ( -1+i \right )t}+\left ( \frac{1}{2}-\frac{i}{2} \right )e^{\left ( -1-i \right )t}$

Which I believe is right, however when I try to show that the curved part of the contour goes to zero as R tends to infinity I find that it does not, and so my answer if not what I want.

Now I have been looking at ways around this and jordans lemma keeps appearing, although I really have no idea how to apply it (or even if this is the right approach), could someone please explain it or point me in the right direction?

Thanks,

Lewis.

Hello Lewis and welcome to MHF.

When you say the standard contour what do you mean? You need to integrate on a vertical line that is to the right of the real part of all of the poles of the integrand.

One such vertical line is

$s=iz$

The imaginary axis.

This gives the integral

$\frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{ize^{izt}}{(iz+1)^2+1}(idz)= \frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{ze^{izt}}{(z-i-1)(z-i+1)}dz$

This is now an integral with z real (an integral on the real axis) edit: sorry this was a bad choice of variables

Now we can apply the residue theorem on the semicircle above the real axis and the line segment on the real axis.

So by the residue theorem we get

$\frac{(i+1)e^{i(i+1)t}}{2}-\frac{(i-1)e^{i(i-1)t}}{2}=-e^{-t}\sin(t)+e^{-t}\cos(t)$

Now the reason we wanted to write the integral this way

$\frac{1}{2\pi i}\int_{-\infty}^{\infty}\frac{ze^{izt}}{(z-i-1)(z-i+1)}dz$

is so we can apply Jordan's lemma

We can identify

$f(z)=\frac{z}{(z-i-1)(z-i+1)}$

and note that

$\lim_{R \to \infty}f(Re^{i\theta})=0$

So by Jordan's Lemma Jordan's Lemma -- from Wolfram MathWorld

$\lim_{R\to \infty}\int e^{itz}f(z)dz=0$

and we are done.
• May 23rd 2011, 02:05 AM
Prove It
Just rewrite this as

$\displaystyle F(s) = \frac{s + 1}{(s + 1)^2 + 1} - \frac{1}{(s + 1)^2 + 1}$

and then use $\displaystyle \mathcal{L}^{-1}\left\{\frac{s - a}{(s - a)^2 + \omega ^2}\right\} = e^{a\,t}\cos{(\omega \,t)}$ and $\displaystyle \mathcal{L}^{-1}\left\{\frac{\omega}{(s - a)^2 + \omega ^2}\right\} = e^{a\,t}\sin{(\omega\,t)}$.
• May 23rd 2011, 11:57 AM
MathMan170
Quote:

Originally Posted by TheEmptySet
Hello Lewis and welcome to MHF.

When you say the standard contour what do you mean? You need to integrate on a vertical line that is to the right of the real part of all of the poles of the integrand.

One such vertical line is

$s=iz$
.
.
.
$\lim_{R \to \infty}f(Re^{i\theta})=0$

So by Jordan's Lemma Jordan's Lemma -- from Wolfram MathWorld

$\lim_{R\to \infty}\int e^{itz}f(z)dz=0$

and we are done.

Thanks alot, it makes alot more sense now, I was just struggling with how I could apply it the integral I had, but I see how now.