# Thread: Integration by Parts (with infinity)

1. ## SOLVED Integration by Parts (with infinity)

$\displaystyle F=\int_0^\infty e^{-st}t^2 dt$

My unsuccessful attempt:
Integration by parts: $\displaystyle \int fg'=fg-\int f'g$
Using "liate":
$\displaystyle f=t^2 f'=2t$
$\displaystyle g=- \frac{1}{s} e^{-st} g'=e^{-st}$

$\displaystyle F=- \frac{1}{s} e^{-st} \cdot t^2 |^\infty_0 - \int _0^\infty 2t\cdot - \frac{1}{s} e^{-st} dt$

$\displaystyle F= \lim_{A \to \infty} (-\frac{1}{s} e^{-sA} \cdot A^2) - \int _0^\infty 2t\cdot - \frac{1}{s} e^{-st} dt$

Using L'Hopital on the limit I get that the limit is equal to 0. So...
$\displaystyle F= \frac{2}{s} \int _0^\infty t\cdot e^{-st} dt$

I use integration by parts again and I get...
$\displaystyle F= \frac{2}{s} [-\frac{1}{s}e^{-st}\cdot t|^\infty _0 -\int _0^\infty e^{-st}dt ]$
$\displaystyle F= \frac{2}{s} [\lim_{A\to\infty} (0) + \frac{1}{s} \lim_{A\to\infty} (-1)]$

$\displaystyle F= \frac{-2}{s^2}$

The answer is supposed to be $\displaystyle F=\frac{2!}{s^3}$

Where did I go wrong?

2. With the substitution $\displaystyle s t= \tau$ You obtain...

$\displaystyle \int _{0}^{\infty} t^{2}\ e^{-s t}\ dt = \frac{1}{s^{3}}\ \int_{0}^{\infty} \tau^{2} e^{-\tau}\ d \tau$ (1)

... where the s variable is 'pulled out' the integral so that integration [by parts...] is more confortable...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. I'm probably making a really dumb mistake but I don't see where the $\displaystyle s^3$ comes from.

$\displaystyle \int _{0}^{\infty} t^{2}\ e^{-s t}\ dt =\int _{0}^{\infty} \frac{\tau ^2}{s^{2}}e^{-\tau}\ d \tau= \frac{1}{s^{2}}\int _{0}^{\infty} \tau ^2 e^{-\tau} d\tau$

4. Originally Posted by Carbon
I'm probably making a really dumb mistake but I don't see where the $\displaystyle s^3$ comes from.

$\displaystyle \int _{0}^{\infty} t^{2}\ e^{-s t}\ dt =\int _{0}^{\infty} \frac{\tau ^2}{s^{2}}e^{-\tau}\ d \tau= \frac{1}{s^{2}}\int _{0}^{\infty} \tau ^2 e^{-\tau} d\tau$
... that is because $\displaystyle s t= \tau \implies dt= \frac{d \tau}{s}$...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. There it is. :P

Thanks! Do you happen to know if this tau substitution works for most cases involving computing laplace transforms?

6. Originally Posted by Carbon
There it is. :P

Thanks! Do you happen to know if this tau substitution works for most cases involving computing laplace transforms?
... pratically the substitution $\displaystyle s t=\tau\ ,\ dt=\frac{d \tau}{s}$ always works...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

7. Your integral = Laplace transform of $\displaystyle t^2$ = $\displaystyle \dfrac{2}{s^3}$.