# A closed form for integrals of the type $\Gamma\left(\sigma+it\right)$?

Numerical evidence strongly suggests that:

$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, \sigma > 0$$

and

$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot (1-\frac11\mathrm{e})\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -1 < \sigma < 0$$

and

$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot \left(1-0\,\mathrm{e}\right)\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -2 < \sigma < -1$$

and

$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot \left(1-\frac12\,\mathrm{e}\right)\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -3 < \sigma < -2$$

and

$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot \left(1-\frac13\,\mathrm{e}\right)\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -4 < \sigma < -3$$

and

$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot \left(1-\frac38\,\mathrm{e}\right)\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -5 < \sigma < -4$$

where the rational factors before $$\mathrm{e}$$ seem to be OEIS: A053557, i.e.:

$$\int_{-\infty}^{\infty}\Gamma\left(\sigma+it\right) \,\mathrm{d}t = 2\cdot \left(1-\mathrm{e}\sum_{k=0}^n \frac{(-1)^k}{k!}\,\right)\cdot\frac{\pi}{\mathrm{e}} \qquad \sigma \in \mathbb{R}, -(n+1) < \sigma < -n, n \in \mathbb{N}$$

I expect this is already known in the math literature, however would like to understand how this could be derived. Tried many sources on the web e.g. the Wiki and Wolfram pages and also OEIS: A061382 however without success. Does anyone know how this is done or would be able to provide a link to a proof?

P.S.: A well-known and maybe related result is:

$$\int_{-\infty}^{\infty}\frac{\cos(x)}{(x^2+1)} \,\mathrm{d}x = 2\cdot\frac{\pi}{\mathrm{e}}$$

• Use the definition of the gamma function to rewrite the integral as a double integral. Commented Jan 7 at 19:42
• You have a Barnes integral. Your sum can be written in terms of the derangement function Commented Jan 7 at 19:52

For $$\Re(s) >0$$, the Mellin transform of $$e^{-x}$$ is $$\Gamma(s)$$.

Therefore, for any value of $$\sigma >0$$, the Mellin inversion theorem states $$e^{-x} = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma+ i \infty} x^{-s} \, \Gamma(s) \, \mathrm ds = \frac{1}{2 \pi} \int_{-\infty}^{\infty} x^{-(\sigma +it)} \, \Gamma(\sigma +it) \, \mathrm dt.$$

Letting $$x=1$$, we get $$e^{-1} = \frac{1}{2 \pi} \int_{-\infty}^{\infty} \Gamma(\sigma +it) \, \mathrm dt , \quad \sigma >0.$$

Now suppose that $$f(x) = \sum_{k=0}^{\infty} \phi(k) (-x)^{k}$$ is a function such that $$\phi$$ satisfies the conditions of Ramanujan's master theorem and $$\int_{0}^{\infty} x^{s-1} \sum_{k=0}^{\infty} \phi(k) (-x)^{k} \mathrm dx = \frac{\pi}{\sin (\pi s)} \, \phi(-s)$$ for $$0 < \Re(s) < \delta$$.

Then for a positive integer $$N$$, $$\int_{0}^{\infty} x^{s-1} \sum_{k=N}^{\infty} \phi(k) (-x)^{k} \, \mathrm dx= \frac{\pi}{\sin (\pi s)} \, \phi(-s)$$ for $$-N < \Re(s) < -N+1$$.

This is Theorem 8.1 in the paper Ramanujan's Master Theorem by Amdeberhan, Espinosa, et al.

The proof given in the paper is to apply Ramanujan's master theorem to the function $$\sum_{k=0}^{\infty} \phi(k+N) (-x)^{k}$$ and then shift the parameter $$s$$.

So for $$-1 < \Re(s) <0$$, the Mellin transform of $$e^{-x}-1= \sum_{n=1}^{\infty} \frac{1}{\Gamma(n+1)} (-x)^{n}$$ is also $$\Gamma(s)$$.

Therefore, for any value of $$\sigma$$ between $$-1$$ and $$0$$, the Mellin inversion theorem states $$e^{-x}-1= \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma+ i \infty} x^{-s} \, \Gamma(s) \, \mathrm ds = \frac{1}{2 \pi} \int_{-\infty}^{\infty} x^{-(\sigma +it)} \, \Gamma(\sigma +it) \, \mathrm dt.$$

Letting $$x=1$$, we have $$e^{-1}-1 = \frac{1}{2\pi} \int_{-\infty}^{\infty} \Gamma(\sigma +it) \, \mathrm dt, \quad -1 < \sigma <0.$$

Similarly, $$e^{-x}-1 + x \bigg|_{x=1} =e^{-1}= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \Gamma(\sigma +it) \, \mathrm dt, \quad -2 < \sigma <-1,$$

$$e^{-x}-1 + x - \frac{x^{2}}{2} \bigg|_{x=1} =e^{-1} - \frac{1}{2}= \frac{1}{2 \pi} \int_{-\infty}^{\infty} \Gamma(\sigma +it) \, \mathrm dt, \quad -3 < \sigma <-2,$$

and so on.

• This is really beautiful. You actually did provide a more comprehensive answer than the one I had already accepted, but trust it is ok to leave it as is.
– Agno
Commented Jan 7 at 23:44

Use the definition of the gamma function to rewrite the integral as a double integral

$$I = \int_{-\infty}^\infty\int_0^\infty e^{\sigma \log r + it \log r - r}\frac{drdt}{r}= \int_{-\infty}^\infty ds\:e^{\sigma s - e^s}\int_{-\infty}^\infty dt\: e^{ist}$$

$$= \int_{-\infty}^\infty 2\pi\delta(s)e^{\sigma s - e^s}ds = \frac{2\pi}{e}$$

This of course assumes $$\sigma>0$$ to use that form of the integral. To get the values between each nonnatural integer you would use alternative integral representations that differ by the factors you found.

• Very nice! I had tried the double integral, but didn't manage to make the third step.
– Agno
Commented Jan 7 at 22:05