4
$\begingroup$

I calculated this integral: $${\int_{0}^{\infty}e^{-cx}x^{n}\arctan(ax)\mathrm{d}x=\frac{n!}{c^{n+1}}\left(\text{Ci}(b)\sum_{k=0}^{n}\sin\left(b-\frac{k\pi}{2}\right)\frac{b^k}{k!}+\left(\frac{\pi}{2}-\text{Si}(b)\right)\sum_{k=0}^{n}\cos\left(b-\frac{k\pi}{2}\right)\frac{b^k}{k!}+\sum_{k=1}^{n}\frac{1}{k}\sum_{j=0}^{k-1}\sum_{l=0}^{n-k}\frac{\sin\left(\pi\cdot\frac{l-j}{2}\right)}{j!l!}b^{j+l}\right)}$$ Where $b:=\dfrac{c}{a}$
I wanted to ask for some help with manipulating the last summations $$\sum_{k=1}^{n}\frac{1}{k}\sum_{j=0}^{k-1}\sum_{l=0}^{n-k}\frac{\sin\left(\pi\cdot\frac{l-j}{2}\right)}{j!l!}b^{j+l}$$ since the only values ​​that sine can have are $1$, $-1$ and $0$ I imagine that at least one of the summations can be removed.
In general I noticed that: $$\sin\left(\frac{l-j}{2}\pi\right)=\begin{cases} 0&\text{if }j-l=2n_0\\ +1&\text{if }j-l=4n_0+1\\ -1&\text{if }j-l=4n_0-1\end{cases}$$

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

$\endgroup$
7
  • $\begingroup$ 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
    Aug 26 at 16:25
  • 1
    $\begingroup$ 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
    Aug 26 at 16:42
  • $\begingroup$ @Aruralreader I was thinking this sums as a sort a double integral above a triangular domain $\endgroup$
    – Math Attack
    Aug 26 at 17:08
  • $\begingroup$ 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
    Aug 27 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
    Aug 27 at 5:21

0

Reset to default

You must log in to answer this question.