integrate by parts:I subscript m ≡ ∫ [e exp x / (e expx) +1)exp 2] [x exp m] dx, limits -∞ to ∞?

Question

The answer must be in the form:

I subscript 2 = 4∫ x dx / (e exp x)+1 Limits:0 to ∞

a)Then by expanding the integrand in powers of e exp-x and integrating term by term, show that I subscript2 can be expressed as infinite series.

b)Then we get the summation of the series to show that

I subscript2 = Л exp2 / 3

Answer 1

In I_2 you have u = x^2, dv = e^x dx / (1 + e^x)^2.

Then du = 2x dx, v = - 1 / (1 + e^x).

u v at +inf. = u v at -inf. = 0

Integrating - v du yields I_2 = integral(x from -inf. to +inf.) (2x dx / (1 + e^x)), which diverges at the lower end. It looks like they mistakenly assumed it was just double the integral from 0 to +inf., but it isn't. That integral, of 4 x dx / (1 + e^x) from 0 to +inf., becomes the integral of the series

4 x e^-x / (e^-x + 1) = 4 x ( e^-x - e^-2x + e^-3x - e^-4x + ... )

Since the integral of x e^-x dx from 0 to +inf. is Gamma(2) = (2 - 1)! = 1! = 1, and so x e^-nx dx ==> (u/n) e^-u (du/n) integrates to 1/(n^2) over that range. So we end up with

A = 4(1 - 1/(2^2) + 1/(3^2) - 1/(4^2) + ...)

Since pi^2 / 6 = 1 + 1/(2^2) + 1/(3^2) + ..., after some manipulation we get

A - 4(pi^2 / 6) = - 2 (1/(2^2) + 1/(4^2) + 1/(6^2) + ...)
A - 4(pi^2 / 6) = -2 ((pi^2 / 6) / 4)
A = 8 pi^2 / 12 - 1 pi^2 / 12 = 7 pi^2 / 12

Answer 2

tell me 1 and 1 = ?.

























































































11. lol