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? 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)  $.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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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.
