# Thread: A hard to crack definite integral

1. ## A hard to crack definite integral

∫ t^2 e^t Cost dt
0

I have been hinted that this integral can be solved using Laplace transform but I don't see any e^ -t in the integrand and if I assume e^t = e^-(-t) then the result become infinity. Please help.

2. Originally Posted by niz

∫ t^2 e^t Cost dt
0

I have been hinted that this integral can be solved using Laplace transform but I don't see any e^ -t in the integrand and if I assume e^t = e^-(-t) then the result become infinity. Please help.
The result of that integral is infinity (the t^2 and exponential terms blow up) I can only assume you copied it down wrong, or that it was a misprint and they meant e^-t

3. $\int t^{2}e^{t}cos(t)dt$

Tabular integration may work well with this.

$+\rightarrow t^{2}\searrow \;\ \;\ \rightarrow e^{t}cos(t)$

$-\rightarrow 2t\searrow \;\ \;\ \rightarrow \frac{e^{t}cos(t)}{2}+\frac{e^{t}sin(t)}{2}$

$+\rightarrow 2\searrow \;\ \;\ \rightarrow \frac{e^{t}sin(t)}{2}$

$-\rightarrow 0 \;\ \;\ \rightarrow \frac{e^{t}sin(t)}{4}-\frac{e^{t}cos(t)}{4}$

Now, add up the signed products of the diagonals.

$\frac{t^{2}e^{t}cos(t)}{2}+\frac{t^{2}e^{t}sin(t)} {2}-te^{t}sin(t)+\frac{e^{t}sin(t)}{2}-\frac{e^{t}cos(t)}{2}$

Here's the graph.

The lower limit gives -1/2.

But, see what happens with the upper limit?. As LostChild stated, if that were $e^{-t}$ then that is another matter.

If we had $\int_{-\infty}^{0}t^{2}e^{t}cos(t)dt$, then we would have a defintive solution.

4. Yes, I think there was a print mistake in the book. Thank a lot for the help.

5. Originally Posted by niz
Assuming, as everyone said that this should be $\int_0^{\infty}t^2e^{-t}\cos(t)dt$ we can use the trick that then our integral is equal to $\Re\int_0^{\infty}t^2e^{(i-1)t}dt$ which is a much easier integral to compute.