# Math Help - inverse Laplace transform by contour integration

1. ## inverse Laplace transform by contour integration

use contour integration to find the inverse Laplace transform of
$\frac{1}{s^3 + a^3}$

i thought i should first split the fraction, so I got
$\frac{1}{s^3 + a^3} = \frac{1}{(s+a)(s^2 - as + a^2)}$
but i tried to use partial fractions to split this fraction and it didn't work!
the inverse Laplace transform formula is
$f(x) = \frac{1}{2 \pi i}\int^{\gamma + i\infty}_{\gamma - i\infty} \frac{1}{s^3 + a^3}e^{sx} ds$
i'm not sure where to go from here!

2. ## Re: inverse Laplace transform by contour integration

Originally Posted by wik_chick88
use contour integration to find the inverse Laplace transform of
$\frac{1}{s^3 + a^3}$

i thought i should first split the fraction, so I got
$\frac{1}{s^3 + a^3} = \frac{1}{(s+a)(s^2 - as + a^2)}$
but i tried to use partial fractions to split this fraction and it didn't work!
the inverse Laplace transform formula is
$f(x) = \frac{1}{2 \pi i}\int^{\gamma + i\infty}_{\gamma - i\infty} \frac{1}{s^3 + a^3}e^{sx} ds$
i'm not sure where to go from here!
Why should you have to use Contour Integration? Partial Fraction should have put this into a form where you could recognise the Inverse Laplace Transform, or find it on your table...

What did you do with your Partial Fractions?

3. ## Re: inverse Laplace transform by contour integration

well when i did partial fractions i got
$\frac{1}{(s + a)(s^2 - as + a^2)} = \frac{A}{s + a} + \frac{Bs + C}{s^2 - as + a^2}$
equating the coefficients i got
$0 = A + B$
$A = -B$

$0 = -Aa + Ba + C$
$0 = -2Aa + C$
$2Aa = C$
$Aa = \frac{C}{2}$

$1 = Aa^2 + Ca$
$1 = \frac{Ca}{2} + \frac{2Ca}{2}$
$1 = \frac{3Ca}{2}$
$C = \frac{2}{3a}$

so $Aa = \frac{C}{2} = \frac{2}{6a} = \frac{1}{3a}$
$A = \frac{1}{3a^2}$

so $B = -A = -\frac{1}{3a^2}$

$so \frac{1}{(s + a)(s^2 - as + a^2)} = \frac{1}{3a^2(s+a)} + \frac{-\frac{1}{3a^2}s+\frac{2}{3a}}{s^2 - as + a^2}$

is this right so far? from here it just seems to get very complicated and messy, and i am still left with the $s^2$!

4. ## Re: inverse Laplace transform by contour integration

Originally Posted by wik_chick88
well when i did partial fractions i got
$\frac{1}{(s + a)(s^2 - as + a^2)} = \frac{A}{s + a} + \frac{Bs + C}{s^2 - as + a^2}$
equating the coefficients i got
$0 = A + B$
$A = -B$

$0 = -Aa + Ba + C$
$0 = -2Aa + C$
$2Aa = C$
$Aa = \frac{C}{2}$

$1 = Aa^2 + Ca$
$1 = \frac{Ca}{2} + \frac{2Ca}{2}$
$1 = \frac{3Ca}{2}$
$C = \frac{2}{3a}$

so $Aa = \frac{C}{2} = \frac{2}{6a} = \frac{1}{3a}$
$A = \frac{1}{3a^2}$

so $B = -A = -\frac{1}{3a^2}$

$so \frac{1}{(s + a)(s^2 - as + a^2)} = \frac{1}{3a^2(s+a)} + \frac{-\frac{1}{3a^2}s+\frac{2}{3a}}{s^2 - as + a^2}$

is this right so far? from here it just seems to get very complicated and messy, and i am still left with the $s^2$!
\displaystyle \begin{align*} \frac{1}{(s + a)(s^2 - as + a^2)} &= \frac{A}{s + a} + \frac{Bs + C}{s^2 - as + a^2} \\ \frac{1}{(s +a)(s^2 - as + a^2)} &= \frac{A(s^2 - as + a^2) + (Bs + C)(s + a)}{(s + a)(s^2 - as + a^2)} \\ 1 &= A(s^2 - as + a^2) + (Bs + C)(s + a) \\ 1 &= As^2 - Aas + Aa^2 + Bs^2 + Bas + Cs + Ca \\ 0s^2 + 0s + 1 &= (A + B)s^2 + (Ba - Aa + C)s + Aa^2 + Ca \end{align*}

So $\displaystyle A + B = 0, Ba - Aa + C = 0, Aa^2 + Ca = 1$

From the first equation, we get $\displaystyle A = -B$, and substituting into the other two equations gives

$\displaystyle Ba + Ba + C = 0, -Ba^2 + Ca = 1$

$\displaystyle 2Ba + C = 0, (-Ba + C)a = 1$

$\displaystyle 3Ba - Ba + C = 0, -Ba + C = \frac{1}{a}$

Substituting the second of these into the first gives

\displaystyle \begin{align*} 3Ba + \frac{1}{a} &= 0 \\ 3Ba &= -\frac{1}{a} \\ B &= -\frac{1}{3a^2} \end{align*}

Substituting into $\displaystyle -Ba + C = \frac{1}{a}$ gives

\displaystyle \begin{align*} -\left(-\frac{1}{3a^2}\right)a + C &= \frac{1}{a} \\ \frac{1}{3a} + C &= \frac{1}{a} \\ C &= \frac{1}{a} - \frac{1}{3a} \\ C &= \frac{2}{3a} \end{align*}

And substituting into $\displaystyle A + B = 0$ we find

\displaystyle \begin{align*} A - \frac{1}{3a^2} &= 0 \\ A &= \frac{1}{3a^2} \end{align*}

Therefore

$\displaystyle \frac{1}{(s + a)(s^2 - as + a^2)} = \frac{\frac{1}{3a^2}}{s + a} + \frac{-\frac{1}{3a^2}s + \frac{2}{3a}}{s^2 - as + a^2} = \frac{1}{3a^2(s + a)} + \frac{2a - s}{3a^2(s^2 - as + a^2)}$

5. ## Re: inverse Laplace transform by contour integration

Originally Posted by wik_chick88
use contour integration to find the inverse Laplace transform of
$\frac{1}{s^3 + a^3}$

i thought i should first split the fraction, so I got
$\frac{1}{s^3 + a^3} = \frac{1}{(s+a)(s^2 - as + a^2)}$
but i tried to use partial fractions to split this fraction and it didn't work!
the inverse Laplace transform formula is
$f(x) = \frac{1}{2 \pi i}\int^{\gamma + i\infty}_{\gamma - i\infty} \frac{1}{s^3 + a^3}e^{sx} ds$
i'm not sure where to go from here!
The procedure for finding the inverse Laplace Tranform via a contour integration was found by the Britisch mathematician Thomas John l'Anson Bromwich about a centiry ago. The details are in...

The Complex inversion formula. Bromwich contour.

Kind regards

$\chi$ $\sigma$