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When I tried to establish the Maclaurin power series expansion of the reciprocal $\frac{1}{\Gamma(z)}$ of the Euler gamma function $\Gamma(z)$, I came across the improper integral $$ I_k=\int_{0}^{\infty}\frac{\sin t}{t}(\ln t)^k\operatorname{d\!} t,\quad k\in\mathbb{N}_0. $$ I believe that the result I am asking for the improper integral $I_k$ existed somewhere in literature. For example, $$ I_0=\frac{\pi}{2}, \quad I_1=-\frac{\pi}{2}\gamma, \quad I_2=\frac{\pi}{24} \bigl(12 \gamma ^2+\pi ^2\bigr), \quad I_3=-\frac{\pi}{8}\bigl[8\zeta(3)+4 \gamma ^3+\gamma\pi^2\bigr], $$ where \begin{equation*} \gamma=\lim_{n\to\infty}\biggl(1+\frac{1}{2}+\frac{1}{3}+\dotsm+\frac{1}{n}-\ln n\biggr) =0.57721 56649 01532 86060\dotsc \end{equation*} is the Euler--Mascheroni constant, whose irrationality or rationality is still unknown, and $\zeta(3)$ denotes the well-known Riemann zeta function $\zeta(z)=\sum_{n=1}^\infty\frac1{n^z}$.

Where to find or how to establish a general explicit or closed-form formula for the above improper integral $I_k$ for $k\in\mathbb{N}_0$? Thanks.

Adding on 4 November 2024: Prove the positivity \begin{equation*} (-1)^{k}I_k>0, \quad k\in\mathbb{N}_0. \end{equation*}

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    $\begingroup$ we typically have one question per post; you may want to post your new question separately. $\endgroup$ Commented Nov 4 at 18:03
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    $\begingroup$ related $\endgroup$ Commented Nov 4 at 20:15
  • $\begingroup$ @SidharthGhoshal Thanks. $\endgroup$
    – qifeng618
    Commented Nov 5 at 11:52

1 Answer 1

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Start from $$I_k=\int_{0}^{\infty}\frac{\sin t}{t}(\ln t)^k\operatorname{d\!} t = \lim_{p\downarrow 0}\frac{d^k}{dp^{k}}\int_{0}^{\infty}\frac{t^p \sin t}{t}\,dt = \lim_{p\downarrow 0}\frac{d^k}{dp^k}\Gamma(p)\sin(\pi p/2)$$ $$\qquad=k!\left[p^{-1}e^{-\gamma p}\sin(\pi p/2)\exp\left(\sum_{n=2}^\infty(-1)^n n^{-1}\zeta(n)p^n\right)\right]_\text{coefficient of $p^k$ term}$$ $$\qquad=\tfrac{1}{2}k!\operatorname{Im}\left[\exp\left((i\pi/2-\gamma)p+\sum_{n=2}^\infty(-1)^n n^{-1}\zeta(n)p^n\right)\right]_\text{coefficient of $p^{k+1}$ term}.$$ This then becomes the known relation between moments and cumulants, so $$I_k=\frac{1}{2(k+1)}\operatorname{Im}\sum_{n=1}^{k+1}B_{k+1,n}(\kappa_1,\kappa_2,\ldots,\kappa_{k+2-n})$$ $$\kappa_1=i\pi/2-\gamma,\;\;\kappa_n=(-1)^n(n-1)!\zeta(n),\;\;n=2,3,\ldots,$$ with $B$ the incomplete Bell polynomial.

This gives, for example, $$I_6=20 \pi \zeta (3)^2+5 \gamma \pi \left(4 \gamma ^2+\pi ^2\right) \zeta (3)+72 \gamma \pi \zeta (5)+\frac{275 \pi ^7}{2688}+\frac{19 \gamma ^2 \pi ^5}{32}+\frac{5 \gamma ^4 \pi ^3}{8}+\frac{\gamma ^6 \pi }{2}$$
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    $\begingroup$ I voted up this answer, but this answer is not the form I would like, because this answer is not explicit and is not to the end. $\endgroup$
    – qifeng618
    Commented Sep 30 at 9:14

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