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  Easy Math Editor

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\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

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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

Pi Han Goh - 4 years, 8 months ago

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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

Uzumaki Nagato Tenshou Uzumaki - 4 years, 8 months ago

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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, ....

Pi Han Goh - 4 years, 8 months ago

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@Pi Han Goh ok :D i very understand it thx you so much :D

Uzumaki Nagato Tenshou Uzumaki - 4 years, 8 months ago

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