# What is the value of $\Gamma(\mathrm{i})$ ?

What is the value of $\Gamma(\mathrm{i})$ ? $\Gamma(z)$ is Gamma function. Here $\mathrm{i}^2=-1$.Can you help me with this problem ?

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en.wikipedia.org/wiki/… – M.B. Jun 4 '12 at 12:26
M.B., why don't you make this comment into an answer? – g.castro Jun 4 '12 at 12:35
Sure, I can do that. – M.B. Jun 4 '12 at 12:41

According to Wikipedia the value is: $\Gamma(i) = (-1+i)! \approx -0.1549 - 0.4980i$.

Now from J.M.'s comment we know that $|\Gamma(i)|^2 = \frac{\pi}{\sinh \pi}$ but I do not think $\Gamma(i)$ can be expressed by elementary functions.

edit: more identities (including the one above) can be found at http://mathworld.wolfram.com/GammaFunction.html

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So people don't have to click on links: there is the identity $|\Gamma(i)|^2=\dfrac{\pi}{\sinh\pi}$, so that takes care of the absolute value at least. The argument, however, doesn't lend itself well to a simple expression... – J. M. Jun 4 '12 at 12:49
@J.M. This is a very nice closed expression, which one will, however, not find when clicking on the wikipedia link, unless I've overlooked it. So do you happen to have a source for this particular expression?. – user20266 Jun 4 '12 at 12:55
@Thomas: under 'additional identities': mathworld.wolfram.com/GammaFunction.html – M.B. Jun 4 '12 at 12:56
@M.B. Thank you very much! – Daoyi Peng Jun 6 '12 at 8:07

So: a method for computing it. The integral formula converges to compute $\Gamma(1+i)$, then the functional equation will give us $\Gamma(i)$ from that. $$\Gamma(i) = \int_{0}^{\infty} \operatorname{e} ^{-x} \operatorname{sin} \bigl(\operatorname{log} (x)\bigr) d x - i \int_{0}^{\infty} \operatorname{e} ^{-x} \operatorname{cos} \bigl(\operatorname{log} (x)\bigr) d x$$

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Thank you very much! – Daoyi Peng Jun 6 '12 at 8:06
Now, I wonder if any computer software or book of integral tables lists $\int_{0}^{\infty} \operatorname{e} ^{-x} \operatorname{sin} \bigl(\operatorname{log} (x)\bigr) d x = \mathrm{Re}\Gamma(i)$ as a "closed form" for that integral? – GEdgar Jun 6 '12 at 13:44