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I've conjectured the following identity for $n\geq0$ integers: $$ \int_0^1 \ln(x)^n \operatorname{Ei}(x) \, dx = (-1)^{n+1}n! \cdot \left(-\operatorname{Ei}(1)+\sum_{k=1}^{n+1} {_kF_k}\left(\begin{array}c1,1,\dots,1\\2,2,\dots,2\end{array}\middle|\,1\right)\right), $$ where $\ln$ is the natural logarithm, $\operatorname{Ei}$ is the exponential integral and ${_pF_q}$ is the generalized hypergeometric function.

For $n=0$ it is $1-e+\operatorname{Ei}(1)$ and for $n=1$ it is $e-1-\gamma$, where $e$ is base of the natural logarithm, and $\gamma$ is the Euler-Mascheroni constant.

How could we prove this identity?

user153012
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  • Seeing this $(-1)^{n+1}n!$, it could probably done by taking the $n$th derivative of $\int_0^1 x^k\operatorname{Ei}(x)dx$ at $k=0$, while using $\operatorname{Ei}(x)=\gamma+\ln x+\sum_{n \ge 1} \frac{x^n}{n n!}$... – Noam Shalev - nospoon Aug 15 '15 at 14:26
  • the cases $n=0,n=1$ are easily calculated using integration by parts. maybe this can be generalized to arbitrary $n$ nice question btw. – tired Aug 15 '15 at 14:28
  • You can go by parts. The obstacle there is to show that $\int_0^1log^n(x)e^x$ yields the hypergeometric functions you mentioned! – tired Aug 15 '15 at 14:33
  • I think a double sum involving Hypergeometric functions might be involved I was working on the integral and that popped out. – Oussama Boussif Aug 26 '15 at 22:06

1 Answers1

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We can do the first two cases by integrating by parts and using the fact that $$\text{Ei}(x) = \ln (x) + \gamma + \mathcal{O}(x) \, , \quad x>0.$$

For the first one, we have

$$ \begin{align} \int_{0}^{1} \text{Ei}(x) \, dx &= x \text{Ei}(x) \Big|^{1}_{0^{+}} - \int_{0}^{1}e^{x} \, dx \\ &= \text{Ei}(1)- e +1. \end{align}$$

And for the second one, we have

$$\begin{align}\int_{0}^{1} \ln(x) \text{Ei}(x) \, dx &= \text{Ei}(x) \left(x\ln (x) -x \right) \Bigg|^{1}_{0^{+}}-\int_{0}^{1} e^{x} \ln (x) \, dx + \int_{0}^{1} e^{x} \, dx \\ &= -\text{Ei}(1) - \Big(e^{x} \ln(x) - \text{Ei}(x) \Big) \Bigg|^{1}_{0^{+}} + e -1 \\ &= - \text{Ei}(1) + \text{Ei}(1) - \gamma + e -1 \\ &= -\gamma + e -1 \end{align}$$

since

$$\lim_{x \to 0^{+}} \Big(e^{x} \ln(x) - \text{Ei}(x) \Big) = \lim_{x \to 0^{+}} \Big(\big(1+\mathcal{O}(x) \big)\ln(x) - \ln(x) - \gamma + \mathcal{O}(x) \Big) = - \gamma.$$

Random Variable
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