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Find $\int e^{x/y} dy$. Now this is part of an incredibly long exercise but I just got stuck in here, and I feel weird because I don't know how to solve this.

nullUser
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mathobs
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    You might be weird, but it is not because you cannot solve this. – Thomas Andrews Jun 24 '13 at 15:03
  • It feels weird because I have solved more difficult integrals,and i cant find this.. – mathobs Jun 24 '13 at 15:06
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    Somehow I connect this exercise with one about double integrals, where the order of integration can make a pretty hard (or even impossible) integral into a rather easy one... – DonAntonio Jun 24 '13 at 15:07
  • According to an old problem I found, this is related to the Exponential Integral. So it does not have a formula in terms of the elementary functions. But that only problem's answer was on Yahoo! Answers, so it might not be reliable. http://answers.yahoo.com/question/index?qid=20100119111548AAegZ0h – Thomas Andrews Jun 24 '13 at 15:13
  • Are you sure the integral is $dy$? Because $dx$ is much easier. :) – Thomas Andrews Jun 24 '13 at 15:13

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In fact we can express $\int e^{\frac{x}{y}}~dy$ as series form:

$\int e^{\frac{x}{y}}~dy=\int\sum\limits_{n=0}^\infty\dfrac{\left(\dfrac{x}{y}\right)^n}{n!}dy=\int\sum\limits_{n=0}^\infty\dfrac{x^ny^{-n}}{n!}dy=\int\left(1+\dfrac{x}{y}+\sum\limits_{n=2}^\infty\dfrac{x^ny^{-n}}{n!}\right)dy=y+x\ln y+\sum\limits_{n=2}^\infty\dfrac{x^ny^{1-n}}{n!(1-n)}+C=y+x\ln y-\sum\limits_{n=2}^\infty\dfrac{x^n}{n!(n-1)y^{n-1}}+C$

Harry Peter
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