Question

`\displaystyle \rm \int_(0)^( \infty ) \sum_(n = 0)^( \infty ) \frac{( - 1 {)}^(n) \: {x}^(2n + 1) }{ {2}^(n) \cdot n!} \sum_(n = 0)^( \infty ) \frac{ {x}^(2n) \: dx }{ {2}^(2n) \cdot(n! {)}^(2) }`​

Answer 1

You can check that both series converge for all real x (say, using the ratio test).

Now,

`\displaystyle e^x = \sum_(n=0)^\infty (x^n)/(n!) \implies x e^(-x^2/2) = \sum_(n=0)^\infty ((-1)^n x^(2n+1))/(2^n n!)`

and if you have some familiarity with differential equations, you might recognize that the second series represents a Bessel function, namely

`\displaystyle I_0(x) = \sum_(n=0)^\infty (x^(2n))/(2^(2n) (n!)^2)`

but we won't bother making use of this.

In the integral, we interchange the integral with the sum

`\displaystyle \int_0^\infty x e^(-x^2/2) \sum_(n=0)^\infty (x^(2n))/(2^(2n) (n!)^2) \, dx = \sum_(n=0)^\infty \frac1{2^(2n) (n!)^2} \int_0^\infty x^(2n+1) e^(-x^2/2) \, dx`

and consider the sequence of integrals,

`J_n = \displaystyle \int_0^\infty x^(2n) x e^(-x^2/2) \, dx`

Integrating by parts with

`u = x^(2n) \implies du = 2n x^(2n-1) \, dx`

`dv = x e^(-x^2/2) \, dx \implies v = -e^(-x^2/2)`

Then the contribution of uv is 0, so

`J_n = \displaystyle 2n \int_0^\infty x^(2(n-1)) x e^(-x^2/2) \, dx = 2n J_(n-1)`

We have

`J_0 = \displaystyle \int_0^\infty x e^(-x^2/2) \, dx = 1`

so that

`J_n = 2n J_(n-1) = 2n (2(n-1)) J_(n-2) = 2n (2(n-1)) (2(n-2)) J_(n-3)`

and so on; after k steps,

`J_n = 2n (2(n-1)) (2(n-2)) \cdots (2(n-(k-1))) J_(n-k)`

so that when n = k,

`J_n = 2n (2(n-1)) (2(n-2)) \cdots 2 J_0 = 2^n n!`

Then the integral we want is

`\displaystyle \int_0^\infty x e^(-x^2/2) \sum_(n=0)^\infty (x^(2n))/(2^(2n) (n!)^2) \, dx = \sum_(n=0)^\infty (2^n n!)/(2^(2n) (n!)^2) = \sum_(n=0)^\infty \frac1{2^n n!}`

and knowing the series expansion for
`e^x`, it follows that the integral has a value of
`e^(1/2) = \boxed{√(e)}`