Question

Integrate this

Answer 1

\[\int\limits_{0}^{\infty} xdx/(e^{x}-1)\]

Answer 2

Don't you think we should use ILATE method here???

Answer 3

what is ILATE method? explain or link me. tnx

Answer 4

ILATE means Inverse Trigonometric, Logarithmic, Algebraic, Trigonometric and Exponential..
It is order wise..

The formula for this goes as:

\[\int\limits_{}^{} uv = u \int\limits_{}^{}v.dx - (\int\limits_{}^{}(du/dx) \times (\int\limits_{}^{}v.dx))).dx\]

Answer 5

oh no that will not help

Answer 6

I know the answer but I wanted to discuss it with you math lovers

Answer 7

Where u and v can be any of the functions..
Here, u is taken as the first function and v is taken as the second function..

Priority order of these function is given by ILATE..

For example : If I would like to solve the integral for : xsinx.

Then I will take x as first function i.e u and sinx as second function ie v..

Ok then discuss my friend..

Answer 8

\[\int\limits_{0}^{\infty} xdx/(e^x-1)=\int\limits_{0}^{\infty} (x e^{-x})/(1-e^{-x})dx\]

Answer 9

Yes you have taken e^x common from the denominator and then taken it in the numerator..
What in next???

Answer 10

\[0<x <\infty \rightarrow 0<e^{-x}<1 \ \ the \ we \ can \ write \\ \\ \\ 1/(1-e^{-x})=1+e^{-x}+e^{-2x}+...\]

Answer 11

then*

Answer 12

series are very useful for such integrals

Answer 13

\[now \ we \ have \ I= \int\limits_{0}^{\infty} (x e^{-x})(\sum_{n=0}^{\infty} \ e^{-nx})dx \\ =\sum_{n=0}^{\infty} \ \int\limits_{0}^{\infty} \ x e^{-(n+1)x}dx \\ = \sum_{n=0}^{\infty} \ 1/(1+n)^{2} \ = \zeta(2)= \pi^{2}/6\]

Answer 14

please use limit.. \[lim_{m rightarrow infty} int_{0}^{m} your equation\]
and solve it.. can you?

Answer 15

i dont understand what is that?lol

Answer 16

do u want it clear than this?

Answer 17

i can explain in more details if u want

Answer 18

that's what i mean