I'm struggling to evaluate the definite integral of $\int_0^1e^x\ln(x)dx$. By integration by parts I get the antiderivative to be $e^x\ln(x)-\operatorname{Ei}(x)$. Im having trouble evaluating the limits of integration. At x=1 I get $-\operatorname{Ei}(1)$ but at x=0 I cannot get a value. Wolfram alpha says the two sided limit as x approaches 0 does not exist but says that the definite integral is $\gamma – \operatorname{Ei}(1)$ and I cannot figure out why. If the limit as x approaches zero does not exist, how can Wolfram get a value for it? It may be possible to transform the integral to have bounds of $0$ to $\infty$ and use a property of the Laplace Transform that says $\int_0^{\infty}fg=\int_0^{\infty}\mathcal{L}(f)\mathcal{L}^{-1}(g)$ but this seems pretty far fetched. Any ideas as to how WFA got it's result?
Evaluate $\int_0^1 e^x \ln(x)dx$
definite integralsintegration
Best Answer
When you evaluate an improper integral where its bound is a singularity, you don't take the limit as $x$ approaches that bound on both sides, as you can see from the definition of improper integrals. In our case, the integral $\displaystyle \int_{0}^{1}e^{x}\ln\left(x\right)dx$ is improperly integrable because $\displaystyle \lim_{a\to 0^+} \int_{a}^{1}e^{x}\ln\left(x\right)dx$ converges.
For this quick calculation, we will use the Puiseux Series of the Exponential Integral
$$\operatorname{Ei}(x) = \gamma + \ln(x) + \sum_{n=1}^{\infty}\frac{x^{n}}{nn!}.$$
Indeed,
$$ \begin{align} \int_{0}^{1}e^{x}\ln\left(x\right)dx &= \lim_{a\to 0^+} \int_{a}^{1}e^{x}\ln\left(x\right)dx \\ &= \lim_{a\to 0^+}\Big[e^x \ln(x) - \operatorname{Ei}(x)\Big]_{a}^{1} \\ &= e^1 \ln(1) - \operatorname{Ei}(1) - \lim_{a\to 0^+} \left(e^a \ln(a) - \operatorname{Ei}(a)\right) \\ &= -\operatorname{Ei}(1) - \lim_{a\to 0^+}\left(e^a \ln(a) - \left(\gamma + \ln(a) + \sum_{n=1}^{\infty}\frac{a^n}{nn!}\right)\right) \\ &= -\operatorname{Ei}(1) - \lim_{a\to 0^+} \left(e^a \ln(a) - \ln(a)\right) + \lim_{a\to 0^+} \sum_{n=1}^{\infty}\frac{a^n}{nn!} + \lim_{a\to 0^+}\gamma \\ &= -\operatorname{Ei}(1) - 0 + 0 + \gamma \\ &= \gamma - \operatorname{Ei}(1). \\ \end{align} $$