How to find the imporper integral from - 00 to 00 of f(x) = (e^(ikx)).(1-e^x / 1+e^x) where k is a real fixed number ??

 

I tried to write e^ikx = coskx + i sinkx but i don't know what to do with the term multiplied by e^ikx which is (1-e^x / 1+e^x) ??

 

 

Thank you

How to find the imporper integral from - 00 to 00 of f(x) = (e^(ikx)).(1-e^x / 1+e^x) where k is a real fixed number ??

 

I tried to write e^ikx = coskx + i sinkx but i don't know what to do with the term multiplied by e^ikx which is (1-e^x / 1+e^x) ??

 

 

Thank you

 

Are you shure the integral is findable?

Yes , that's what the professor said.

Isn't correct that i-e^x / 1+e^x = - tanh(x/2) ???

 

If yes, then how can we proceed ??

It may be easier to work with if you first rearrange the fraction:

 

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

 

Since the integral of this latter fraction is more easily calculable, and is equal to

 

x - 2ln(1+e^x)

I submitted it to WolframAlpha http://www.wolframalpha.com/input/?i=improper+integral+calculator

for k = 0 and 1. It could find an antiderivative, but the integral diverges.

 

Perhaps hdg7833 have been given the solution by his professor, now, and he can share it with us ?

How to find the imporper integral from - 00 to 00 of f(x) = (e^(ikx)).(1-e^x / 1+e^x) where k is a real fixed number ??

 

I tried to write e^ikx = coskx + i sinkx but i don't know what to do with the term multiplied by e^ikx which is (1-e^x / 1+e^x) ??

 

 

Thank you

I'm confused. Is there something I'm missing? Is that a limited of integration?

I obtain this solution as an indefinite integral:

 

[latex]e^{\frac{ik}{2}}\left(2e^{\frac{1}{2}}-1\right)u + 2\left(1-e^{\frac{1}{2}}\right)ln|u| + C[/latex]

Edited January 11, 201312 yr by x(x-y)