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Context

I calculated this integral: $$\begin{array}{l} \displaystyle\int_{0}^{\infty}e^{-cx}x^{n}\arctan(ax)\mathrm{d}x=\\ \displaystyle\frac{n!}{c^{n+1}}\left\lbrace\sum_{k=0}^{n}\left[\text{Ci}\left(b\right)\sin\left(b-\frac{k\pi}{2}\right)+\left(\frac{\pi}{2}-\text{Si}\left(b\right)\right)\cos\left(b-\frac{k\pi}{2}\right)\right]\frac{b^{k}}{k!}\right.\\ \displaystyle\left.\quad\qquad+\sum_{k=1}^{n}\frac{1}{k}\sum_{j=1}^{k}\sum_{l=0}^{n-k}\frac{\cos\left(\frac{\pi\left(l-j\right)}{2}\right)}{\left(j-1\right)!l!}b^{j+l-1}\right\rbrace\end{array}$$ Where:

  • $b:=\dfrac{c}{a}$
  • $\text{Si}(z)$ is the sine integral function
  • $\text{Ci}(z)$ is the cosine integral function

Question

I need help simplifying the last two summations: $$\sum_{j=1}^{k}\sum_{l=0}^{n-k}\frac{\cos\left(\frac{\pi\left(l-j\right)}{2}\right)}{\left(j-1\right)!l!}b^{j+l-1}$$

I can't find the transformation that allows me to calculate the sum only on the even "$l-j$" given that: $$\cos\left(\frac{\pi\left(l-j\right)}{2}\right)=\begin{cases}(-1)^{l-j}&\text{if }l-j\text{ is even}\\ 0&\text{if }l-j\text{ is odd}\end{cases}$$

My idea
I noticed that the polynomial is odd, so technically it is possible to substitute $l-j=2s$ where $1\leq s\leq \left\lfloor\dfrac{n}{2}\right\rfloor$?
Thanks in advance for any suggestions

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  • $\begingroup$ If you visualize the entries of the $l,j$ summation in a $n-k+1\times k$ array, all of the terms in the square of size $\min(n-k+1,k)$ cancel due to odd symmetry. This only leaves you with the largest terms in the summation, in the rectangular strip leftover (if there is one). $\endgroup$
    – Ninad Munshi
    Commented Aug 26, 2023 at 16:25
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    $\begingroup$ You might explore writing the $1/j!l!$ term as ${j+l\choose j}/(j+l)!$, the sine term in terms of exponentials, and introduce the change of variable $m = j+l$. $\endgroup$
    – A rural reader
    Commented Aug 26, 2023 at 16:42
  • $\begingroup$ @Aruralreader I was thinking this sums as a sort a double integral above a triangular domain $\endgroup$
    – Math Attack
    Commented Aug 26, 2023 at 17:08
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    $\begingroup$ For whatever it's worth, Mathematica gives the integral as $\frac{2^{n-1} \text{sgn}(a) G_{2,4}^{4,1}\left(\frac{c^2}{4 a^2}|\begin{array}{c} \frac{1}{2},1 \\ 0,\frac{1}{2},\frac{n+1}{2},\frac{n+2}{2} \\\end{array}\right)}{\sqrt{\pi } c^{n+1}}$ where $G$ is the Meijer G function. $\endgroup$
    – JimB
    Commented Aug 27, 2023 at 1:53
  • $\begingroup$ The most inner sum is already a monster : beside a bunch of floor function appear four generalized hypergeometric functions $\endgroup$
    – Claude Leibovici
    Commented Aug 27, 2023 at 5:21

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