Difficult integral: $\int e^x \ln (x)\, \mathrm dx$

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?

• – Lucian Feb 12 '14 at 21:44

$$\int e^x\ln xdx=e^x\ln x-\int\frac{e^x}xdx=e^x\ln x-\text{Ei}(x).$$
• That should be $+\text{Ei}(x)$, not $-\text{Ei}(x)$. – John Moeller Feb 12 '14 at 22:07
• @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. – Lucian Feb 12 '14 at 22:22
• $\displaystyle{\large{\rm E_{i}}\left(x\right)}$ 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 $\displaystyle{\large x\not=0}$. – Felix Marin Feb 12 '14 at 22:43
• @FelixMarin: All indefinite integrals are definite ones; e.g. $\displaystyle\int e^xdx=\int_{-\infty}^xe^tdt$. Similarly, $\displaystyle\int\frac{e^x}xdx=$ $=\displaystyle\int_{-\infty}^x\frac{e^t}tdt=\text{Ei(x)}$. – Lucian Feb 12 '14 at 23:24