Help with the integral $\int_{0}^{\infty}\frac{x^{s}}{\Gamma(s)}s^{z-1}ds$

Consider the integral :

$$\int_{0}^{\infty}\frac{x^{s}}{\Gamma(s)}s^{z-1}ds$$ Where $$x\in \mathbb{R}^{+}$$ , $$z\in \mathbb{C}$$

Using the Hankel contour representation of the reciprocal gamma function, we have:

$$\int_{0}^{\infty}\frac{x^{s}}{\Gamma(s)}s^{z-1}ds=\frac{i}{2\pi}\oint \int_{0}^{\infty} \left(-\frac{x}{t}\right)^{s}s^{z-1}e^{-t}dsdt$$ $$=\Gamma(z)\frac{i}{2\pi}\oint \left(-\log\left(-\frac{x}{t}\right)\right)^{-z}e^{-t}dt$$

But i have no idea on how to do this integral. any help is highly appreciated

• Who said it has a simpler expression ? You have found its analytic continuation. The last integral is entire and for integer $-z$ it is the derivatives of $\oint (-t)^{-z} e^{-xt}d(xt) = (1-e^{-2i \pi z})(-1)^{-z} x^z \Gamma(z)$ (the latter with the reflection formula gives the contour representation for $1/\Gamma(1-z)$) Commented Oct 1, 2019 at 11:34
• Note that the original integral only converges for $\Re(z) > -1$. Commented Oct 1, 2019 at 12:32

It looks like Ramanujan's master theorem (see [1] pg. 298-300). Instead of Ramanujan's formula we use Hardy's theorem which is also in [1] pg. 299-300.

THEOREM 1. (Ramanujan-Hardy) Let $$s=\sigma+it$$, $$\sigma,t$$ both real. Let $$H(\delta)=\{s:\sigma\geq-\delta\}$$, $$0<\delta<1$$. If $$\psi(s)$$ is analytic on $$H(\delta)$$ and exist constants $$C,P,A$$, with $$A<\pi$$ such that $$|\psi(s)|\leq Ce^{P\sigma+A|t|}\textrm{, }\forall s\in H(\delta),$$ for $$x>0$$ and $$0, we define $$\Psi(x)=\frac{1}{2\pi i}\int^{c+i\infty}_{c-i\infty}\frac{\pi}{\sin(\pi s)}\psi(-s)x^{-s}ds.$$ If $$0, then $$\Psi(x)=\sum^{\infty}_{k=0}\psi(k)(-x)^k.$$ For $$0<\sigma<\delta$$, we have $$\int^{\infty}_{0}\Psi(x)x^{s-1}dx=\frac{\pi}{\sin(\pi s)}\psi(-s).$$

Here (in your case) if we set $$\psi(x)=\frac{\phi(x)}{\Gamma(x+1)},$$ and $$\Psi(x)=\frac{a^x}{\Gamma(x)}=\sum^{\infty}_{k=0}\psi(k)(-x)^k=\sum^{\infty}_{k=0}\frac{\phi(k)}{k!}(-x)^k,$$ then $$\int^{\infty}_{0}\frac{a^x}{\Gamma(x)}x^{s-1}dx=\frac{\pi}{\sin(\pi s)}\psi(-s)=\frac{\pi}{\sin(\pi s)}\frac{\phi(-s)}{\Gamma(1-s)}$$ Using now the formula $$\frac{\pi}{\sin(\pi s)\Gamma(1-s)}=\Gamma(s),$$ we arrive to $$I(s):=\int^{\infty}_{0}\frac{a^x}{\Gamma(x)}x^{s-1}dx=\Gamma(s)\phi(-s).$$ Hence if $$(Mf)(s)$$ is the Mellin transform of $$f$$, then $$\left(Mf\right)(s)=\int^{\infty}_{0}f(x)x^{s-1}dx$$ and $$\left(M\Psi\right)(s)=I(s)=\int^{\infty}_{0}\frac{a^x}{\Gamma(x)}x^{s-1}dx =\Gamma(s)\phi(-s).$$ But from 2 we have the following

THEOREM 2. (for the conditions see 2) $$\int^{\infty}_{-\infty}\left(M\Psi\right)(\sigma+it)f(t)dt=2\pi\sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}f(i(\sigma+k)).$$

Hence in our case with $$\Psi(t)=a^t/\Gamma(t)$$ and $$f(t)=e^{-itx}(M\Psi)(\sigma-it)$$, we have $$\int^{\infty}_{-\infty}(M\Psi)(\sigma+it)f(t)dt=$$ $$=\int^{\infty}_{-\infty}(M\Psi)(\sigma+it)e^{-itx}(M\Psi)(\sigma-it)dt=$$ $$=2\pi\sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}f(i(\sigma+k))=$$ $$=2\pi\sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}e^{-ii(\sigma+k)x}\Gamma(\sigma-ii(\sigma+k))\phi(-\sigma+ii(\sigma+k))=$$ $$=2\pi\sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}e^{(\sigma+k)x}\Gamma(2\sigma+k)\phi(-2\sigma-k).$$ Hence with $$\sigma=1/2$$: $$\int^{\infty}_{-\infty}\left|(M\Psi)\left(\frac{1}{2}+it\right)\right|^2e^{-itx}dt=2\pi e^{x/2} \sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}\Gamma(k+1)\phi(-k-1)e^{kx}.$$ Hence we can write $$\int^{\infty}_{-\infty}\left|(M\Psi)\left(\frac{1}{2}+it\right)\right|^2e^{itx}dt=2\pi e^{-x/2} \sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}\left(\int^{\infty}_{0}\frac{a^t}{\Gamma(t)}t^{k}dt\right)e^{-kx}\textrm{, }x>0$$ Which is a sampling formula, recovering the absolute value of the Mellin transform of $$\Psi(x)$$, from the values of $$(M\Psi)(x)$$ at $$x=k+1$$, where $$k$$ belongs to the non negative integers $$k=0,1,2,\ldots$$. As someone can see the result can be generalized very easily for $$\Psi(x)$$ analyitic around $$0$$ and entire in $$\textbf{C}$$ such that $$\int^{\infty}_{0}|\Psi(t)t^{k}|dt<\infty$$.

NOTES.

1) The general formula that rises is $$\int^{\infty}_{-\infty}\left|(M\Psi)\left(\frac{1}{2}+it\right)\right|^2e^{itx}dt=2\pi e^{-x/2} \sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}(M\Psi)(k+1)e^{-kx}\textrm{, }x>0$$ As an example of evaluation take $$\Psi(x)=e^{-x}$$. Then $$\int^{\infty}_{0}e^{-t}t^{k}dt=k!.$$ Hence we get the next integral $$\int^{\infty}_{-\infty}\left|\Gamma\left(\frac{1}{2}+it\right)\right|^2e^{itx}dt=2\pi\frac{e^{x/2}}{e^x+1}\textrm{, }x>0.$$

2)

Also from the inverse Fourier theorem with $$\Psi(x)=\frac{a^x}{\Gamma(x)}$$, then $$\left|(M\Psi)\left(\frac{1}{2}+iw\right)\right|^2=\left|\int^{\infty}_{0}\frac{a^t}{\Gamma(t)}t^{-1/2+iw}dt\right|^2=$$ $$=\int^{\infty}_{-\infty}e^{-x/2} \sum^{\infty}_{k=0}\frac{\Psi^{(k)}(0)}{k!}\left(\int^{\infty}_{0}\frac{a^t}{\Gamma(t)}t^{k}dt\right)e^{-kx}e^{-i x w}dx$$

[1]: Bruce. C. Berndt. "Ramanujan's Notebooks Part 1". Springer-Verlang. New York, Berlin, Heidelberg, Tokyo. 1985.

2: N.D. Bagis."Numerical Evaluations of Functions Series and Integral Transforms with New Sampling Methods". Thesis. Aristotele University of Thessaloniki, Greece (2007), (in Greek from Researchgate here)

• Thanks for the effort. But i am a bit at loss. how does one use the formulas above to evaluate the integral in question ? Commented Oct 2, 2019 at 8:39
• You asking for ($M\Psi)(s)$, $s\in\textbf{C}$. I have prove that evaluating $\left|M\Psi(1/2+iy)\right|$, $y\in\textbf{R}$, reduces to know only the values of $(M\Psi)(k+1)=\int^{\infty}_{0}a^t t^k/\Gamma(t)dt$, $k\in\textbf{Z}_{+}$. Commented Oct 2, 2019 at 16:29
• I have no idea of what you are doing Nikos, $f(x)=\frac{a^{x-1}}{\Gamma(x)}$ is analytic and fast decreasing thus its Mellin transform is $\phi(s)\Gamma(s)$ with $\phi(s)$ entire and $\phi(-k) = (-1)^k f^{(k)}(0)$. Then what ? $\sum_{k\ge 0} \frac{\phi(-k)(-1)^k / k!}{s+k}$ is the Mellin transform of $f(x)1_{x< 1}$ not of $f$. Commented Oct 2, 2019 at 20:34
• Let $\Psi(x)$ be analytic around 0, with radious of convergence $r>0$ and $a$ such that $0<a<r$. Let also $x_0\in\textbf{R}$ such that $\int^{\infty}_{0}|\Psi(u)|u^{x_0-1}du<\infty$. Then $\int^{a}_{0}\Psi(u)u^{z-1}du=\sum^{\infty}_{m=0}\frac{\Psi^{(m)}(0)}{m!}\frac{a^{z+m}}{z+m}$ is meromorphic at $\textbf{C}$ with simple poles at $z=-m$, $m\in\textbf{Z}_{+}$. Also then $\int^{\infty}_{a}\Psi(u)u^{z-1}du$ is analytic when $Re(z)<x_0$. Hence we can use residual calculus in the meromorphic function $f(z)(M\Psi)(x+i z)$ to get Theorem 2 (under some convergence conditions). See [2]. Commented Oct 3, 2019 at 4:06