I arrived at the following result

$$\tag{1}\int^\infty_0 z^{p-1} E^2(z)\,dz=\frac{\Gamma(p)}{p}\int^1_0\frac{_2F_1(p,p;p+1;-\frac{1}{z})}{z}\,dz$$

where the exponential integral $E(z)$ is defined as

$$E(z)=\int^\infty_z \frac{e^{-t}}{t}\,dt$$

I have two questions

[1] Does (1) hold for all $p>0$ ?

[2] Is there a way to simplify or solve the integral on the right ?

  • $\begingroup$ The 2F1 and Gamma functions hold for $p > 0$. The problem is the lower limit of the integral. The integral involves calculating $\int_{0}^{1} \frac{dy}{y^{r+1}}$ $\endgroup$ – Leucippus May 20 '14 at 18:56
  • $\begingroup$ I think the right-hand side of your equation is undefined. As @Leucippus have written in the primitive function at $z=0$ you run into a dividing by zero. $\endgroup$ – user153012 Sep 24 '14 at 21:04
  • $\begingroup$ @user153012 Perhaps I'm missing something, but I'm fairly sure the RHS is defined for all $p>0$. In particular, we can verify that the RHS is $\ln{4}$ at $p=1$. $\endgroup$ – David H Dec 24 '14 at 17:45

Let \begin{align} E_{1}(x) = \int_{x}^{\infty} \frac{e^{-t}}{t} \, dt \end{align} then the following two identities can be seen to be \begin{align} \int_{0}^{\infty} x^{n} E_{1}(ax) E_{1}(bx) \, dx &= - \frac{n!}{n+1} \left[ \frac{1}{a^{n+1}} \left\{ \ln\left(\frac{b}{a+b}\right) + \sum_{m=1}^{n} \frac{1}{m} \left( \frac{a}{a+b} \right)^{m} \right\} \right. \\ & \hspace{20mm} \left. + \frac{1}{b^{n+1}} \left\{ \ln\left(\frac{a}{a+b}\right) + \sum_{m=1}^{n} \frac{1}{m} \left( \frac{b}{a+b} \right)^{m} \right\} \right]. \end{align} When $a=b=1$ this becomes \begin{align} \int_{0}^{\infty} x^{p} E_{1}^{2}(x) \, dx &= \frac{\Gamma(p+1)}{2^{p} \, (p+1)} \, \sum_{m=0}^{\infty} \frac{1}{2^{m} (m+p+1)} . \end{align}


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