Evaluate  int_(-oo)^(oo) \ x^2e^(-4x) \ dx ?

Nov 2, 2017

The integral is divergent.

Explanation:

We seek:

${\int}_{- \infty}^{\infty} \setminus {x}^{2} {e}^{- 4 x} \setminus \mathrm{dx}$

Using common sense analysis we can answer the question without actually performing the integral.

If we apply Integration By Parts twice we will get a solution of the definite integral of the form:

$\int \setminus {x}^{2} {e}^{- 4 x} \setminus \mathrm{dx} = \left(A {x}^{2} + B x + C\right) {e}^{- 4 x} + C$

Thuis we will find that:

${\int}_{- \infty}^{\infty} \setminus {x}^{2} {e}^{- 4 x} \setminus \mathrm{dx} = {\lim}_{N \rightarrow \infty} {\left[\left(A {x}^{2} + B x + C\right) {e}^{- 4 x}\right]}_{- N}^{N}$

And will will get convergence at the upper limit as ${e}^{- \infty} \rightarrow 0$ but divergence at the lower limit as ${e}^{\infty} \rightarrow 0$,

Hence the integral is divergent.

Note:

Two applications of IBP yields:

$\int \setminus {x}^{2} {e}^{- 4 x} \setminus \mathrm{dx} = - \frac{\left(8 {x}^{2} + 4 x + 1\right) {e}^{- 4 x}}{32} + C$