How to do integral ∫ − ∞ ∞ x 4 e − x 2 / 2 d x

Question

How to do integral       ∫          −      ∞              ∞            x    4        e          −              x        2                    /            2        d  x

Travis Sellers · Accepted Answer

Recall that  I  (  a  )  =      ∫          −      ∞              ∞            e          −      a              x        2              d  x  =                    π        a                  I    ′    (  a  )  =      ∫          −      ∞              ∞        (  −      x    2    )      e          −      a              x        2              d  x  =  −            1      2                          π                    a                  3                      /                    2                          I    ″    (  a  )  =      ∫          −      ∞              ∞            x    4        e          −      a              x        2              d  x  =            1      2                  3      2                          π                    a                  5                      /                    2                    Setting a=1, we get      ∫          −      ∞              ∞            x    4        e          −              x        2              d  x  =                    3                  π                    4      From this you get that, in general,      ∫          −      ∞              ∞            x          n            e          −              x        2              d  x  =      {                            0                          If           n           is odd.                                                                1              2                                ⋅                                    3              2                                ⋅                                    5              2                                ⋯                                                    n                −                1                            2                                            π                    =                                                    n                !                                  π                                                                              2                  n                                (                n                                  /                                2                )                !                                                              If           n           is even.

Winston Todd · Accepted Answer

We can use integration by parts twice to get

\begin{aligned} \int_{- \infty}^{\infty} x^{4} e^{- x^{2} / 2} d x & = - \int_{- \infty}^{\infty} x^{3} d e^{- x^{2} / 2} \\ = {[- x^{3} e^{- x^{2} / 2}]}_{- \infty}^{+ \infty} + 3 \int_{- \infty}^{\infty} x^{2} e^{- x^{2} / 2} d x \\ = - 3 \int_{- \infty}^{\infty} x d e^{- x^{2} / 2} \\ = {[- 3 x e^{- x^{2} / 2}]}_{- \infty}^{+ \infty} + 3 \int_{- \infty}^{\infty} e^{- x^{2} / 2} d x \\ = 3 \sqrt{2 π} \end{aligned}

To evaluate the last integral above, we can set

I = \int_{- \infty}^{\infty} e^{- x^{2} / 2} d x

Then convert from rectangular to polar coordinates

\begin{aligned} I^{2} & = \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- (x^{2} + y^{2}) / 2} d x d y \\ = \int_{0}^{2 π} \int_{0}^{\infty} e^{- r^{2} / 2} r d r d θ \\ = - \int_{0}^{2 π} \int_{0}^{\infty} d e^{- r^{2} / 2} d θ \\ = \int_{0}^{2 π} 1 d θ \\ = 2 π \end{aligned}

How to do integral int_(-infty)^infty x^4e^(-x^2/2)dx

Answered question

Answer & Explanation

New Questions in Calculus 1