# Integral infinity

Note by Uzumaki Nagato Tenshou Uzumaki
5 years, 3 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Let S denote the value

Since the integrand is an even function

S = 2 * integrate(0 to infinity) x^2 e^x /(1 + e^x)^2 dx

Apply integration by part, let u = x^2 => du = 2x dx, dv = e^x /(1 + e^x)^2 => v = -1/(1 + e^x)

S/2 = uv - integrate v du

S/2 = -x^2 /(1 + e^x) + 2 integrate(0 to infinity) x/(1 + e^x) dx

Apply the limits from 0 to infinity for -x^2/(1 + e^x), you get 0

S/4 = integrate(0 to infinity) x/(1 + e^x) dx, divide top and bottom by e^x

S/4 = integrate(0 to infinity) xe^(-x) /(1 - (-e^(-x)) dx, this is in the form of a/(1-r), sum of a geometric series

S/4 = integrate(0 to infinity) [ xe^(-x) - xe^(-2x) + xe^(-3x) - xe^(-4x) + ... ] dx

Notice each term follows a Gamma Distribution with alpha = 2, beta = 1/1, 1/2, 1/3, ....

It simplfies to

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

Right hand side is a form of Riemann Zeta function of 2

S/4 = pi^2 /12

S = pi^2 /3

Note that (1/1)^2 - (1/2)^2 + (1/3)^2 - (1/4)^2 + ...

= [ (1/1)^2 + (1/2)^2 + (1/3)^2 + (1/4)^2 + ... ] - 2 [ (1/2)^2 + (1/4)^2 + (1/6)^2 + ... ]

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

= (1/2) * [ (1/1)^2 + (1/2)^2 + (1/3)^2 + (1/4)^2 + ... ]

= (1/2) * pi^2 /6 = pi^2 /12

- 5 years, 3 months ago

may you explain why it can be this.

Notice each term follows a Gamma Distribution with alpha = 2, beta = 1/1, 1/2, 1/3, ....

thx before :D

- 5 years, 3 months ago

integrate(0 to infinity) [ xe^(-x) - xe^(-2x) + xe^(-3x) - xe^(-4x) + ... ] dx

= integrate(0 to infinity) [ xe^(-x) ] dx + integrate(0 to infinity) [ xe^(-2x) ] dx + integrate(0 to infinity) [ xe^(-3x) ] dx + integrate(0 to infinity) [ xe^(-4x) ] dx + integrate(0 to infinity) [ xe^(-4x) ] dx + ...

All the integrand are in the form of a gamma function of x^(alpha - 1) * e^(-x/beta)

So alpha = 2 for all integrals, beta = 1/1, 1/2, 1/3, 1/4, ....

- 5 years, 3 months ago

ok :D i very understand it thx you so much :D

- 5 years, 3 months ago