hgd7833 Posted April 14, 2012 Posted April 14, 2012 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
Anvar Posted April 14, 2012 Posted April 14, 2012 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?
hgd7833 Posted April 15, 2012 Author Posted April 15, 2012 Isn't correct that i-e^x / 1+e^x = - tanh(x/2) ??? If yes, then how can we proceed ??
QuantumBullet Posted December 18, 2012 Posted December 18, 2012 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)
caKus Posted December 21, 2012 Posted December 21, 2012 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 ?
stardustbrain Posted January 7, 2013 Posted January 7, 2013 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?
x(x-y) Posted January 11, 2013 Posted January 11, 2013 (edited) 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, 2013 by x(x-y)
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