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

Question

1616 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-11-02T22:14:34

The integral is divergent.

We seek:

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

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:

$int \ x^2e^(-4x) \ dx = (Ax^2+Bx+C)e^(-4x) + C$

Thuis we will find that:

$int_(-oo)^(oo) \ x^2e^(-4x) \ dx = lim_(N rarr oo)[(Ax^2+Bx+C)e^(-4x)]_(-N)^(N)$

And will will get convergence at the upper limit as $e^(-oo) rarr 0$ but divergence at the lower limit as $e^(oo) rarr 0$ ,

Hence the integral is divergent.

Note:

Two applications of IBP yields:

$int \ x^2e^(-4x) \ dx = -((8x^2+4x+1)e^(-4x))/32 + C$

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