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Improper Integral of the following integral?

  1. Feb 28, 2013 #1
    How would one go about computing the following improper integral, with limits of integration [0,∞) using residues?

    [itex]\int exp(x+1/x)/x[/itex]
  2. jcsd
  3. Feb 28, 2013 #2


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    The integral doesn't converge. There's no point in trying to find a way to prove that using residues.
  4. Feb 28, 2013 #3
    How does it not converge though? I've been confused by this integral for quite some time. Using some crappy software of mine, the integral seems to converge to .2278. However, when I apply residues to this improper integral and integrate along a contour with a counter-clockwise semi-circle in the upper half plane and a small circle with radius approaching 0 around the essential singularity at z=0, I get that the integral is equal to 0, by using the fact that the series representation of this function can be written as the product of two absolutely convergent sums, thus allowing one to use the binomial theorem to write out each summand and then rearrange the terms of the series to arrive at the result of the coefficient of 1/z being equal to the infinite series from n=0 to infinity of a_k, where a_k = kth term of the series = (x^k/k!)^2. Sorry about being messy, I'm kind of in a rush at the moment and don't have time to write out all the appropriate syntax.

  5. Feb 28, 2013 #4


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    For positive ##x##, ##\exp(x + 1/x) \geq \exp(x)##, so the integrand is bounded below by ##e^x/x##. The latter function diverges to infinity as ##x \rightarrow \infty##, so certainly it can't have finite area.
  6. Feb 28, 2013 #5
    Oh WOW! I'm sorry guys. The integrand should be exp[-(x+1/x)]/x, heh.
    Last edited: Mar 1, 2013
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