Gamma Function on negative Fractions

[sahil_time]

Gamma Function on negative Fractions!

If we take a look at the Gamma Function and evaluate the integral by parts then we will get infinity in the first step of Integration by Parts eg:
Integral e^-1*x^-5/3 Limits being 0 to Infinity as usual! If we try to integrate this we will get Infinity??What is the Contradiction here?

 

[mathman]

From wikipedia equation may be in Latex, but that part didn't copy)

Although the Gamma function is defined for all complex numbers except the non-positive integers, it is defined via an improper integral that converges only for complex numbers with a positive real part:

\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,{\rm d}t\,.

This integral function is extended by analytic continuation to all complex numbers except the non-positive integers (where the function has simple poles), yielding the meromorphic function we call the Gamma function.

 

[sahil_time]

From wikipedia equation may be in Latex, but that part didn't copy)

Although the Gamma function is defined for all complex numbers except the non-positive integers, it is defined via an improper integral that converges only for complex numbers with a positive real part:

\Gamma(z) = \int_0^\infty t^{z-1} e^{-t}\,{\rm d}t\,.

This integral function is extended by analytic continuation to all complex numbers except the non-positive integers (where the function has simple poles), yielding the meromorphic function we call the Gamma function.

Is there any other way to explain the answer to my Question?

 

[uart]

Is there any other way to explain the answer to my Question?

Yes. Mathman's explanation is correct, but in slightly simpler terms the analytic extension of the Gamma function to Re(z)<0 is made via the recursion relationship.

[tex]\Gamma(n+1) = n \Gamma(n)[/tex]

Or in the reverse direction,

[tex]\Gamma(n-1) = \frac{\Gamma(n)}{n-1}[/tex]

So for example,

[tex]\Gamma(-\frac{1}{2}) = -2 \, \Gamma(\frac{1}{2})[/tex]

and,

[tex]\Gamma(-\frac{3}{2}) = \frac{4}{3} \, \Gamma(\frac{1}{2})[/tex]

etc.

 

[sahil_time]

Yes. Mathman's explanation is correct, but in slightly simpler terms the analytic extension of the Gamma function to Re(z)<0 is made via the recursion relationship.

[tex]\Gamma(n+1) = n \Gamma(n)[/tex]

Or in the reverse direction,

[tex]\Gamma(n-1) = \frac{\Gamma(n)}{n-1}[/tex]

So for example,

[tex]\Gamma(-\frac{1}{2}) = -2 \, \Gamma(\frac{1}{2})[/tex]

and,


[tex]\Gamma(-\frac{3}{2}) = \frac{4}{3} \, \Gamma(\frac{1}{2})[/tex]

etc.

Yeah I agree with u but If u Integrate (e^-x*x^-5/3) by parts taking limits as 0 to Infinity then u won't get a determinate value coz it will give u Infinity!:/

 

[mathman]

You seem to have missed the point! The integral defines the gamma function only for the arguments z, where Re(z) > 0. Otherwise the integral diverges, as you have noticed. However you can get the definition of Gamma function for other values, except negative integers, by using the recursive relation: Γ(z)=Γ(z+1)/z.

 

[uart]

Yeah I agree with u but If u Integrate (e^-x*x^-5/3) by parts taking limits as 0 to Infinity then u won't get a determinate value coz it will give u Infinity!:/

Well that's why they call it analytic extension.

Let me give you another ultra simple example of domain extension that I bet you've been using for years without a second thought.

Almost certainly you would have originally learned that the "Sine" function, [itex]\sin(\theta)[/itex], is defined as the ratio opposite to hypotenuse in a right angle triangle. However there's a problem that you can't even construct a right angle triangle with an angle greater than 90 degrees, so it follows that the domain of the "sine" function is only 0 to 90 degrees.

I'm sure that's not the domain you currently use for "sine" though. So somewhere along your maths journey you extended the domain, can you remember how that happened? You found another working definition for sine that, while consistent with the original definition on the limited domain (acute angles), could be extended over the full circle. That new definition of course being the y coordinate of the position [itex]1\, \angle \theta[/itex] on the unit circle.

 

[sahil_time]

You seem to have missed the point! The integral defines the gamma function only for the arguments z, where Re(z) > 0. Otherwise the integral diverges, as you have noticed. However you can get the definition of Gamma function for other values, except negative integers, by using the recursive relation: Γ(z)=Γ(z+1)/z.

Yeah Then For all the negative fractions and integers it must diverge?isnt t?

 

[uart]

Yeah Then For all the negative fractions and integers it must diverge?isnt t?

It diverges for non positive integers, but I don't see why you think there's a problem with negative fractions. Can you elaborate a bit more on what think the problem is?

 

[sahil_time]

It diverges for non positive integers, but I don't see why you think there's a problem with negative fractions. Can you elaborate a bit more on what think the problem is?

What will you get when u integrate (e^-x)*(x^-5/4) By parts
Limits being 0 and infinity!

 

[uart]

What will you get when u integrate (e^-x)*(x^-5/4) By parts
Limits being 0 and infinity!

What to you get for the ratio of opposite to hypotenuse when you draw a right angle triangle where one of the other angles equals 135 degrees. You can't do it right. So the sine of 135 degrees doesn't exist? Is that right, or is there another way?

You just have to read the answers that have already been provided above in this thread.

 

[sahil_time]

Its still Hard to believe!

 

[nonequilibrium]

It's weird that people like you can be in college

 

[mathman]

What will you get when u integrate (e^-x)*(x^-5/4) By parts
Limits being 0 and infinity!

Stop trying to use the integral for Re(z) ≤ 0! You can use the recursion formula. Since Γ(0)= Γ(1)/0, Γ(n) is infinite for integers n ≤ 0. However the extension formula gives finite values for Γ(z) otherwise, for Re(z) ≤ 0.

 

[sahil_time]

It's weird that people like you can be in college

w/e!:/

 

[sahil_time]

Stop trying to use the integral for Re(z) ≤ 0! You can use the recursion formula. Since Γ(0)= Γ(1)/0, Γ(n) is infinite for integers n ≤ 0. However the extension formula gives finite values for Γ(z) otherwise, for Re(z) ≤ 0.

Yeah Thanx!