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$\mathcal{L}\{f(t)\} = \int_0^\infty{e^{-st}t^2}dt$

Integrating by parts:

$u = t^2$

$du = 2tdt$

$v = -\frac{1}{s}e^{-st}$

$dv = e^{-st}dt$

$\int_0^\infty{e^{-st}t^2}dt = -\frac{t^2}{s}e^{-st} + \frac{2}{s}\int_0^\infty{e^{-st}tdt}$

Integrating by parts on $\int_0^\infty{e^{-st}t}dt$:

$u = t$

$du = dt$

$v = -\frac{1}{s}e^{-st}$

$dv = e^{-st}dt$

$\int_0^\infty{e^{-st}t}dt = -\frac{t}{s}e^{-st}+\frac{1}{s}\int_0^\infty{e^{-st}dt}$

$\int_0^\infty{e^{-st}t}dt = -\frac{t}{s}e^{-st}+\frac{1}{s^2}$

In total:

$\int_0^\infty{e^{-st}t^2}dt = \frac{t^2}{s}e^{-st}+\frac{2}{s}[-\frac{t}{s}e^{-st}+\frac{1}{s^2}]$

I know this is incorrect, but I can't figure out where I'm messing up. Can someone help?

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$$\int_0^\infty{e^{-st}t}dt = -\frac{t}{s}e^{-st}+\frac{1}{s}\int_0^\infty{e^{-st}dt}$$ You forgot a negative, it should instead be $$\int_0^\infty{e^{-st}t}dt = -\frac{t}{s}e^{-st}-\frac{1}{s^2}$$ and also dont forget to take the limits. –  Q.matin Dec 9 '12 at 6:58
1  
Why not just substitute $u = st$ in the original integral? –  Antonio Vargas Dec 9 '12 at 7:17

1 Answer 1

up vote 1 down vote accepted

The integrals are over limits. Just substitute limits for the first term of the integration by parts term and you are through.

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So my final step should be to take the limit for the final integral, evaluating it at b (which goes to infinity) and 0? –  Bailor Tow Dec 9 '12 at 7:00
    
@BailorTow yes, take the limits but also look at my comment above. You forgot a negative and an exponential on $$\frac{1}{s}\int_0^\infty{e^{-st}dt}$$ it should equal $$\int_0^\infty{e^{-st}t}dt = -\frac{t}{s}e^{-st}-\frac{1}{s^2}e^{-st}$$ –  Q.matin Dec 9 '12 at 7:07

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