I am trying to solve an integral:

$\int e^x \ln(x)\,\mathrm dx =\ ?$

I have tried integration by parts and I found out that this method doesn't provide a solution in this case? How to solve it then?


$$\int e^x\ln xdx=e^x\ln x-\int\frac{e^x}xdx=e^x\ln x-\text{Ei}(x).$$

  • $\begingroup$ That should be $+\text{Ei}(x)$, not $-\text{Ei}(x)$. $\endgroup$ – John Moeller Feb 12 '14 at 22:07
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    $\begingroup$ @JohnMoeller: Both Mathematica and Wolfram Alpha seem to agree with my result. If you let $x\to-t$, then the two minus signs will cancel each other out. $\endgroup$ – Lucian Feb 12 '14 at 22:22
  • $\begingroup$ Hm, I'm having a hard time trusting Ei as an indefinite integral here. But sure, I'll give Alpha some leeway. $\endgroup$ – John Moeller Feb 12 '14 at 22:25
  • $\begingroup$ $\displaystyle{\large{\rm E_{i}}\left(x\right)}$ must be a definite Integral: $\displaystyle{\large{\rm E_{i}}\left(x\right) = {\rm P.V.}\int_{-\infty}^{x}{{\rm e}^{t} \over t}\,{\rm d}t}$ with $\displaystyle{\large x\not=0}$. $\endgroup$ – Felix Marin Feb 12 '14 at 22:43
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    $\begingroup$ @FelixMarin: All indefinite integrals are definite ones; e.g. $\displaystyle\int e^xdx=\int_{-\infty}^xe^tdt$. Similarly, $\displaystyle\int\frac{e^x}xdx=$ $=\displaystyle\int_{-\infty}^x\frac{e^t}tdt=\text{Ei(x)}$. $\endgroup$ – Lucian Feb 12 '14 at 23:24

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