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I want to simplify the Laplace transform expression of $E_{i}(-y)^{2}$, where $E_{i}(y)$ is the exponential integral defined by $E_{i}(y) = -\int\limits_{-y}^{\infty} \frac{e^{-t}}{t} dt$.

Question: Can someone please show a simpler expression than what I derived, as shown below?

I know that the Laplace Transform $\mathcal{L}_{p} (E_{i}(-y)) = -\frac{log(p+1)}{p}, ROC \in Re(p) \gt 0$

For $E_{i}(-y)^{2}$, I used the formula for laplace transform of product from Wikipedia.

$\mathcal{L}_{p} (E_{i}(-y)^{2}) = \frac{1}{2\pi i} \int\limits_{c-i \infty}^{c + i \infty} \left(-\frac{log(1+\sigma)}{\sigma}\right) \left(-\frac{log(1+p-\sigma)}{p-\sigma}\right) d\sigma$, where $c$ is a vertical line right of origin (to be in ROC of one of these).

I closed the contour on the left half of the complex plane with a keyhole contour, with the hole at $(-1,0)$. If I am right, the integrals for big semicircle and the small circle at the hole vanish. Then I am left with two lines in the contour: one from $-\infty$ to $-1$ (slightly above the x-axis) and one from $-1$ to $-\infty$ slightly below x-axis.

Doing some math (not entirely sure of its correctness), I got the integral to be

$$-\int\limits_{0}^{\infty} \frac{log(p+2+x)}{(p+1+x)(x+1)} dx$$

When I entered this in Wolfram, it gave me this complicated result:

enter image description here

As you can see, the tool didn't simplify it. I am not very familiar with Polylog functions. Is there a simplification possible?

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  • $\begingroup$ In the remaining 2 segments of the contour I chose, they become $2\pi i$. $\endgroup$
    – Srini
    Commented Apr 27 at 19:19
  • $\begingroup$ Now my comment looks out of context. I can swear Mark Viola had asked a question, but I can't see it now :) $\endgroup$
    – Srini
    Commented Apr 27 at 19:20
  • $\begingroup$ Anyone else that has the question of how 2 log terms became one, like the more mysteriously disappeared question, see example 5 here: en.wikipedia.org/wiki/Contour_integration . My key hole is at -1 instead of at origin, otherwise the same principle $\endgroup$
    – Srini
    Commented Apr 27 at 22:36
  • $\begingroup$ Your integral for the Laplace transform is correct except there shouldn't be a negative sign out front: $F(p)=\int\limits_{0}^{\infty} \frac{log(p+2+x)}{(p+1+x)(x+1)} dx$. I'll try to write a full answer when I get the time. Cheers :) $\endgroup$
    – David H
    Commented Jun 22 at 2:43
  • $\begingroup$ Thanks David. Now that you have navigated your way to this question, I owe you a full explanation of why I made you do those tedious double summations (for which I am grateful). Let me write up the whole (naive) story, when I get time. $\endgroup$
    – Srini
    Commented Jun 22 at 16:25

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