solve this partial integration with step function

Question

I would like to know how to solve this partial integration. The equation I got is based on the following convolution:

$$t^2e^{-2t} * te^t$$

The part I am having a hard time with is the (t-u) coefficient.

$$\int_0^t ue^{u} (t-u)e^{(t-u)}du$$ $$ = \int_0^t ue^{u} e^t e^{-u} (t-u)du$$ $$ = \int_0^t ue^{-2u} (t-u)e^{t}du$$

I tried turning the convolution the other way around but I think this might result in an even harder integration.

Answer 1

I would do the following way :

Let, $ t-u = p ==> dt = dp $

$\int_{-u}^0 p e^p dp$

Now try integration by parts.

$\int_{-u}^0 p e^p dp = pe^p|_{-u}^0 - \int_{-u}^0 e^p dp = -e^{-u} - 1+e^{-u}$

The answer is:

$-e^{p-t} - 1+e^{p-t} $

Hope this might help!

Answer 2

Since the question is tagged Laplace transform, I suspect that the convolution should be $$ t^2e^{-2t}*te^t=\int_0^t u^2e^{-2u}(t-u)e^{t-u}\,du. $$ Using linearity (and the fact that the $t$-dependence can go outside the integral), this integral can be written as $$ te^t\int_0^tu^2e^{-3u}\,du-e^t\int_0^tu^3e^{-3u}\,du. $$ I suggest that you start to integrate by parts in both integrals (differentiating the polynomial part of course). Tell me if you have a problem doing so.

Answer 3

You have edited the question:

Ok now you may try the following way :

$\int (tue^{tu}) - \int u^2 e^t$

then use integration by parts for the first part and the second part is easy integration.