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 ?
What is the value of $\Gamma(\mathrm{i})$ ?
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0Sure, I can do that. – 2012-06-04
2 Answers
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.
Wikipedia: http://en.wikipedia.org/wiki/Particular_values_of_the_Gamma_function#Imaginary_unit
edit: more identities (including the one above) can be found at http://mathworld.wolfram.com/GammaFunction.html
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0@M.B. Thank you very much! – 2012-06-06
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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1Now, 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? – 2012-06-06