The formula $\int_0^{\infty} t^{k} {\tt sech}(t)dt=\frac{(-1)^k}{2^{2k+1}} \left( \psi^{(k) } \left( \frac {3} {4} \right) -\psi^{(k)}\left( \frac {1} {4} \right) \right) $ is interesting; however, it is only true whenever k is a nonnegative integer. Does there exist a formula for $\int_0^{\infty} t^{k} {\tt sech}(t)dt$ that is correct whenever the real part of k is greater than negative 1?
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$\bf{\int_0^{\infty} t^{k-1} f(t)dt=(-1)^k \left[ F^{(k-1)}(s) \right]_{s=0}^{s=\infty}}$.
Consider the Laplace transform of f(t) to be F(s), given by F(s)=$\int_0^{\infty} f(t) e^{-st} dt$. We have that the Laplace transform of $t^n f(t)$ is $(-1)^n F^{(n)}(s)$, and we have that $\int_0^{\infty} \frac {f(t)} {t} dt = \int_0^{\infty} F(s)ds$. Hence, $\int_0^{\infty} \frac {t^k f(t)} {t} dt = \int_0^{\infty} (-1)^k F^{(k)}(s) ds$. This can be rewritten as $\int_0^{\infty} t^{k-1} f(t) dt = (-1)^k \int_0^{\infty} F^{(k)}(s) ds $; furthermore, $\int_0^{\infty} t^{k-1} f(t)dt=(-1)^k \left[ F^{(k-1)}(s) \right]_{s=0}^{s=\infty}$.
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The Laplace transform of ${\tt sech}$(t) is $\bf{\frac {1} {2}\left( \psi^{(0) } \left( \frac {s+3} {4} \right) -\psi^{(0)}\left( \frac {s+1} {4} \right) \right) }$.
In the next section, we must know the Laplace transform of ${\tt sech}(t)$. Consider the fact that ${\tt sech}(t)=\frac {2} {e^t+e^{-t}}$. Then, $F(s)=\int_0^{\infty} \frac {2} {e^t+e^{-t}} e^{-st} dt$. Consider the digamma function, $\psi^{(0)}(z)=\int_0^{\infty} \frac {e^{-t}} {t}-\frac {e^{=zt}} {1-e^{-t}} dt$. $\int_0^{\infty} \frac{2} {e^t+e^-t}e^{-st}dt$ = $2 \int_0^\infty \frac {e^t} {e^{2t}+1} e^{-st}dt$ = $\frac {1} {2} \int_0^\infty \frac {e^{\frac {1} {4} t}} {e^{\frac {t} {2}}+1} e^{-\frac{1}{4}st}dt$ = $\frac {1} {2} \int_0^{\infty} \frac { e^{\frac{1} {4}(t-st)}} {e^{ \frac{t} {2} }+1}dt$ = $\frac {1} {2} \int_0^{\infty} \frac { e^{t-\frac{1} {4}(s+3)t}} {e^{ \frac{t} {2} }+1}dt$ = $\frac {1} {2} \int_0^{\infty} \frac {e^{-\frac{s+1}{4}t}-e^{-\frac{s+3}{4}t}} {1-e^{-t}}dt$ = $\frac{1}{2}\int_0^{\infty} \frac {e^{-\frac{s+1} {4}t}} {1-e^{-t}}-\frac{e^{-\frac{s+3}{4}t}}{1-e^{-t}}dt$ = $\frac {1} {2}\left( \int_0^{\infty} \frac {e^{-t}} {t} - \frac {e^{-\frac {s+3} {4}t}} {1-e^{-t}}dt - \int_0^{\infty} \frac {e^{-t}} {t} - \frac {e^{-\frac {s+1} {4}t}} {1-e^{-t}}dt \right) $ = $\frac {1} {2}\left( \psi^{(0) } \left( \frac {s+3} {4} \right) -\psi^{(0)}\left( \frac {s+1} {4} \right) \right) $
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$\bf{\int_0^{\infty} t^{k} {\tt sech}(t)dt=(-1)^k \frac{1}{2^{2k+1}} \left( \psi^{(k) } \left( \frac {3} {4} \right) -\psi^{(k)}\left( \frac {1} {4} \right) \right) }$
Consider $f(t)={\tt sech}(t)$. Then, we have $\int_0^{\infty} t^{k-1} {\tt sech}(t)dt=(-1)^k \left[ \frac{1}{2}\frac{d^{k-1}} {ds^{k-1} } \left( \psi^{(0) } \left( \frac {s+3} {4} \right) -\psi^{(0)}\left( \frac {s+1} {4} \right) \right) \right]_{s=0}^{s=\infty}$. Hence, $\int_0^{\infty} t^{k-1} {\tt sech}(t)dt=(-1)^k \frac{1}{2^{2k-2}} \left[ \frac{1}{2} \left( \psi^{(k-1) } \left( \frac {s+3} {4} \right) -\psi^{(k-1)}\left( \frac {s+1} {4} \right) \right) \right]_{s=0}^{s=\infty}$. First, we take the limit at infinity: $\lim_{s\rightarrow\infty} \frac{1}{2}\left( \psi^{(k-1) } \left( \frac {s+3} {4} \right) -\psi^{(k-1)}\left( \frac {s+1} {4} \right) \right) = \lim_{s\rightarrow\infty} \int_{0}^{\infty} \frac {2} {e^t+e^{-t}}e^{-st}dt$, which is zero. Thus, we have: $\int_0^{\infty} t^{k-1} {\tt sech}(t)dt=-(-1)^k \frac{1}{2^{2k-1}} \left( \psi^{(k-1) } \left( \frac {3} {4} \right) -\psi^{(k-1)}\left( \frac {1} {4} \right) \right) $, which can be alternatively written as $\int_0^{\infty} t^{k} {\tt sech}(t)dt=(-1)^k \frac{1}{2^{2k+1}} \left( \psi^{(k) } \left( \frac {3} {4} \right) -\psi^{(k)}\left( \frac {1} {4} \right) \right) $.
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ The answer you are looking for is related to the Euler Number $\ds{E_{n}}$ because $$ \on{sech}\pars{x} = \sum_{n = 0}^{\infty}{E_{2n} \over \pars{2n}!}\,x^{2n} $$ I was trying to use the Ramanujan's Master Theorem by rewriting the above expression as $$ \on{sech}\pars{\root{x}} = \sum_{n = 0}^{\infty} \color{red}{{\Gamma\pars{1 + n}\cos\pars{n\pi} \over \Gamma\pars{1 + 2n}}\,E_{2n}} \,{\pars{-x}^{n} \over n!} $$ For this purpose, it was obvious that we need a $\ds{E_{\nu}}$ analytical continuation. Indeed, I found a paper where the author claims he found the coveted continuation -as related to the Riemman $\ds{\zeta}$ function-. However, I was unable to reproduce his results. My "checking" didn't agree with his claim. I hope this "big comment/no answer" will be helpful to you.