# Math Help - Gamma(i)

1. ## 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.

2. 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: $\Gamma (p) = \int_{0}^{\infty} e^{-x}x^{p-1}~dx$ ? that's just the definition though

3. Originally Posted by Jhevon
do you mean: $\Gamma (x) = \int_{0}^{\infty} e^{-x}x^{p-1}~dx$ ? that's just the definition though
(It should be $\Gamma(p)$).
Yes, but the beautiful Gamma function is defined for complex numbers with real part positive as well.

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

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.

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

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

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

5. ## 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.

6. 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.

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

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

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

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

What next?

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