Convergence of Integral: How to Prove for 0<k<1?

rioo
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Homework Statement


Show that \int^{\infty}_{-\infty} \frac{e^{kx}}{1+e^{x}}dx converges if 0&lt;k&lt;1


Homework Equations


None


The Attempt at a Solution


Well if I can show that the integral is dominated by another that converges then I'm done, but I haven't been able to come up with one. I've tried manipulating the integrand (moving the e^{kx} to the bottom and checking limits. The integrand does go to zero at -\infty \mathrm{and\ } \infty, but that doesn't guarantee convergence...
 
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rioo said:

Homework Statement


Show that \int^{\infty}_{-\infty} \frac{e^{kx}}{1+e^{x}}dx converges if 0&lt;k&lt;1


Homework Equations


None


The Attempt at a Solution


Well if I can show that the integral is dominated by another that converges then I'm done, but I haven't been able to come up with one. I've tried manipulating the integrand (moving the e^{kx} to the bottom and checking limits. The integrand does go to zero at -\infty \mathrm{and\ } \infty, but that doesn't guarantee convergence...

Look at the two cases ##\int_0^\infty## and ##\int_{-\infty}^0## separately and use different overestimates on the different intervals. If you can show they are both finite you are done.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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