# Gamma(i)

• Sep 24th 2007, 12:05 AM
bobbyk
Gamma(i)
Express Gamma(i) in terms of elementary functions, if possible.

(i is the imaginary unit and Gamma is Euler's Gamma Function)
(Elementary functions are as defined in Wickepedia)

Bob.
• Sep 24th 2007, 03:43 AM
Jhevon
Quote:

Originally Posted by bobbyk
Express Gamma(i) in terms of elementary functions, if possible.

(i is the imaginary unit and Gamma is Euler's Gamma Function)
(Elementary functions are as defined in Wickepedia)

Bob.

do you mean: $\displaystyle \Gamma (p) = \int_{0}^{\infty} e^{-x}x^{p-1}~dx$ ? that's just the definition though
• Sep 24th 2007, 10:00 AM
ThePerfectHacker
Quote:

Originally Posted by Jhevon
do you mean: $\displaystyle \Gamma (x) = \int_{0}^{\infty} e^{-x}x^{p-1}~dx$ ? that's just the definition though

(It should be $\displaystyle \Gamma(p)$).
Yes, but the beautiful Gamma function is defined for complex numbers with real part positive as well.

$\displaystyle \Gamma(i) = \int_0^{\infty} e^{-z}z^{1-i} dz$

But I am afraid I cannot do this. I do not even want to try. MathWorld has so many identities and non involving imaginary numbers.
• Sep 24th 2007, 01:24 PM
topsquark
Quote:

Originally Posted by ThePerfectHacker
(It should be $\displaystyle \Gamma(p)$).
Yes, but the beautiful Gamma function is defined for complex numbers with real part positive as well.

$\displaystyle \Gamma(i) = \int_0^{\infty} e^{-z}z^{1-i} dz$

But I am afraid I cannot do this. I do not even want to try. MathWorld has so many identities and non involving imaginary numbers.

Shouldn't the contour integration of this function should be the same as in the real case? I'm not even going to try it, but that might be a possible way to evaluate it.

-Dan
• Sep 24th 2007, 01:47 PM
bobbyk
Gamma(i)
Thanks, all!

I have the numerical value of Gamma(i). Its modulus is an elementary
function: Sqrt(pi/sinh(pi)).
So its real and imaginary parts COULD be elementary functions too, but
I haven't been able to find them.

Bob.
• Sep 24th 2007, 04:57 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Shouldn't the contour integration of this function should be the same as in the real case? I'm not even going to try it, but that might be a possible way to evaluate it.

-Dan

I am not exactly sure what you mean. (Wondering)
• Sep 24th 2007, 05:05 PM
ThePerfectHacker
$\displaystyle \Gamma (i) = \int_0^{\infty} e^{-z} z^{i-1} dz$

Now, $\displaystyle z^{i} = e^{\log(z)i} = \cos (\log z)+i\sin (\log z)$

Thus,
$\displaystyle \Gamma (i) = \int_0^{\infty} z^{-1}e^{-z}\cos (\log z) dz + i\int_0^{\infty} z^{-1}e^{-z} \sin (\log z) dz$

So we end up with,
$\displaystyle \int_{-\infty}^{\infty} e^{-e^x} \cos x dx + i \int_{-\infty}^{\infty} e^{-e^x} \sin x dx$

What next?
• Sep 25th 2007, 05:47 AM
topsquark
Quote:

Originally Posted by ThePerfectHacker
I am not exactly sure what you mean. (Wondering)

(sigh) I can't find a picture of the contour on the net. :( But the gamma function may be integrated using complex integration. I was wondering if it could be used for this problem.

-Dan