# Gamma function

• Nov 18th 2008, 10:58 PM
namelessguy
Gamma function
I am working on this problem below.
Prove that $\displaystyle \mid \Gamma(1/2+it) \mid=\sqrt{\frac{2\pi}{e^{\pi t}+e^{-\pi t}}}$, $\displaystyle t \in R$
So, I tried to compute $\displaystyle {\mid \Gamma(1/2+it) \mid}^2$, which I think is equal to $\displaystyle \mid \Gamma(1/2+it) \overline{\Gamma(1/2+it) \mid}$
But here I need to show that $\displaystyle \overline{\Gamma(1/2+it)}=\Gamma(1/2-it)$, and I'm stuck on showing this. Can anyone give some help?
• Nov 19th 2008, 03:33 AM
David24
Quote:

Originally Posted by namelessguy
I am working on this problem below.
Prove that $\displaystyle \mid \Gamma(1/2+it) \mid=\sqrt{\frac{2\pi}{e^{\pi t}+e^{-\pi t}}}$, $\displaystyle t \in R$
So, I tried to compute $\displaystyle {\mid \Gamma(1/2+it) \mid}^2$, which I think is equal to $\displaystyle \mid \Gamma(1/2+it) \overline{\Gamma(1/2+it) \mid}$
But here I need to show that $\displaystyle \overline{\Gamma(1/2+it)}=\Gamma(1/2-it)$, and I'm stuck on showing this. Can anyone give some help?

This should point you in the right direction,
Gamma function - Wikipedia, the free encyclopedia
please let me know if you require any further assistance

Great problem!

David
• Nov 20th 2008, 01:37 AM
namelessguy
Thanks for your help David.
I already showed that
$\displaystyle \mid \Gamma(1/2+it) \overline{\Gamma(1/2-it)} \mid=\frac{2\pi}{e^{\pi t}+e^{-\pi t}}$by a direct computation using the fact that $\displaystyle \Gamma(x)\Gamma(1-x)=\frac{\pi}{sin(\pi x)}$. So, just take the absolute value of both sides then I got the result.
However,I'm still stuck with proving my original question. From the link you gave me, there are lots of ways to define the gamma function, so I think I should use the standard definition
$\displaystyle \Gamma(z)=\int_{0}^{\infty} t^{z-1}e^{-t}dt$
Then I want to show $\displaystyle \overline{\Gamma(z)}=\Gamma\overline{(z)}$,
I have $\displaystyle \overline{\Gamma(z)}=\overline{\int_{0}^{\infty} t^{z-1}e^{-t}dt}=\int_{0}^{\infty}\overline{t^{z-1}e^{-t}dt}$
How do I go from here, anyone can give a hand is appreciated.
• Nov 20th 2008, 01:47 AM
David24
Quote:

Originally Posted by namelessguy
Thanks for your help David.
I already showed that
$\displaystyle \mid \Gamma(1/2+it) \overline{\Gamma(1/2-it)} \mid=\frac{2\pi}{e^{\pi t}+e^{-\pi t}}$by a direct computation using the fact that $\displaystyle \Gamma(x)\Gamma(1-x)=\frac{\pi}{sin(\pi x)}$. So, just take the absolute value of both sides then I got the result.
However,I'm still stuck with proving my original question. From the link you gave me, there are lots of ways to define the gamma function, so I think I should use the standard definition
$\displaystyle \Gamma(z)=\int_{0}^{\infty} t^{z-1}e^{-t}dt$
Then I want to show $\displaystyle \overline{\Gamma(z)}=\Gamma\overline{(z)}$,
I have $\displaystyle \overline{\Gamma(z)}=\overline{\int_{0}^{\infty} t^{z-1}e^{-t}dt}=\int_{0}^{\infty}\overline{t^{z-1}e^{-t}dt}$
How do I go from here, anyone can give a hand is appreciated.

Ahh you obviously didn't read the post in too much detail !! (j/j), there is a fantastic property of the gamma function in terms evaluating the conjugate of F(z) Whereby
Conjugate(F(z)) = F(Conjugate of z)
(I can prove if you want (more than happy to) but its a lot of work...)

Hope this points you in the right direction, if you need any further help please message me - this is a great problem!!

Regards,

David
• Nov 20th 2008, 06:59 AM
namelessguy
Quote:

Originally Posted by David24
Ahh you obviously didn't read the post in too much detail !! (j/j), there is a fantastic property of the gamma function in terms evaluating the conjugate of F(z) Whereby
Conjugate(F(z)) = F(Conjugate of z)
(I can prove if you want (more than happy to) but its a lot of work...)

Hope this points you in the right direction, if you need any further help please message me - this is a great problem!!

Regards,

David

Well, actually I did read that article in details, and I did see the property. But I wanted to prove it before using it. I guess I'm just going to cheat(Giggle) use it by mentioning the property without proving it in this problem. Thanks for the help.
• Nov 20th 2008, 08:08 AM
David24
Quote:

Originally Posted by namelessguy
Well, actually I did read that article in details, and I did see the property. But I wanted to prove it before using it. I guess I'm just going to cheat(Giggle) use it by mentioning the property without proving it in this problem. Thanks for the help.

NamelessGuy,

I apologise if I caused any offence, It was not my intention to do so, I only commented on the 'Conjugate' property as I thought you may not have read it.

Please let me know how you got the solution out, I am highly interested in comparing methodologies.

Regards,

David
• Nov 21st 2008, 12:22 AM
David24
hey mate, have you got it out yet? cant wait to here the result!
• Nov 21st 2008, 01:18 AM
namelessguy
Quote:

Originally Posted by David24
hey mate, have you got it out yet? cant wait to here the result!

Nope, you didn't offense me at all. I'm here to learn and get help so any help is welcome and appreciate.
If you're asking about the solution I got for this. Yes, I did but I didn't prove the conjugation property of the gamma function.
This is how I did
$\displaystyle \mid \Gamma(1/2+it) \overline{\Gamma(1/2+it) \mid}=\mid \Gamma(1/2+it) \Gamma(1/2-it) \mid$
We know that $\displaystyle \Gamma(x)\Gamma(1-x)=\frac{\pi}{sin(\pi x)}$
Hence,$\displaystyle \mid \Gamma(1/2+it) \Gamma(1/2-it) \mid =\mid \frac{\pi}{sin \pi (1/2+it)} \mid$
Now compute $\displaystyle \mid sin \pi (1/2+it) \mid=\mid \frac {e^{i \pi(1/2+it)}-e^{-i \pi(1/2+it)}}{2i} \mid$
$\displaystyle =\mid \frac{ e^{\frac{\pi i}{2}} \times e^{-it} - e^{\frac{-\pi i}{2}} \times e^{it}}{2i} \mid$
$\displaystyle =\frac{ e^{-it} + e^{it}}{2i}$ (here I skipped the step where I evaluated $\displaystyle e^{\frac{\pi i}{2}}$ and $\displaystyle e^{\frac{-\pi i}{2}}$ by using Euler's formula.
So $\displaystyle \mid \Gamma(1/2+it) \Gamma(1/2-it) \mid=\frac{2 \pi}{e^{-it}+e^{it}}$, and we obtain the result by taking square root.