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How to find $$\mathcal{L^{-1}} \left[ \frac{F(s)}{s+a} \right]$$where $F(s)$ is the Laplace transform of $f(t).$

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The title is seriously misleading. –  Did Oct 15 '12 at 15:51
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If you know about convolution, this is just a piece of cake.

$\mathcal L^{-1}\left\{\dfrac{F(s)}{s+a}\right\}$

$=\mathcal L^{-1}\left\{\dfrac{1}{s+a}\right\}*\mathcal L^{-1}\{F(s)\}$

$=e^{-at}*f(t)$

$=\int_0^te^{-a(t-\tau)}f(\tau)~d\tau$

$=e^{-at}\int_0^te^{a\tau}f(\tau)~d\tau$

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Hint:

You know that $$\mathcal{L}(f*g)=F(s)G(s)$$ so $$\mathcal{L^{-1}}\big(F(s)G(s)\big)=f*g$$ wherein $f*g=\int_0^tf(\kappa)g(t-\kappa)d\kappa$.

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Great hint! + 1 –  amWhy Feb 12 '13 at 0:05
    
@Thanks. I hope I do well for to day at M.S.E. –  B. S. Feb 12 '13 at 2:49
    
You're doing well for MSE! –  amWhy Feb 12 '13 at 2:57
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Use laplace transform properties : $$ G(s)=\frac{F(s)}{s+a} $$ $$ \mathcal{L^{-1}}[G(s)]=g(t) $$ $$ \mathcal{L} \left[ e^{at}g(t) \right] = G(s-a)=\frac{F(s-a)}{s} $$ $$ \frac{1}{s} \triangleq \int_0^t $$ $$ e^{at}g(t)=\int_0^t e^{a \tau }f( \tau ) d \tau $$ $$ g(t)=e^{-at} \int_0^t e^{a \tau }f( \tau ) d \tau $$ or we can write : $$ g(t)= \int_0^t e^{-a(t- \tau) }f( \tau ) d \tau $$

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Nice approach +1. –  B. S. Dec 22 '12 at 18:11
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