Gamma Functions

  • #1
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Consider [itex]I(c)= \int_0^\infty \frac{x^{1-\epsilon}}{x+c} dx[/itex]

For what values of [itex]\epsilon[/itex] are these integrals convergent?

Would it be [itex]\epsilon \geq 0[/itex]?

Then I'm asked to use [itex]x^{-\lambda} \Gamma(\lambda) = \int_0^\infty d \alpha \alpha^{\lambda-1}e^{-\alpha x}[/itex] for [itex]x>0[/itex] and the identity [itex]\Gamma(1-\epsilon) \Gamma(\epsilon)=\frac{\pi}{\sin{\pi \epsilon}}[/itex] to show

[itex]I(c)=-\frac{\pi}{\sin{\pi \epsilon}}c^{1-\epsilon}[/itex]

I am really struggling to rewrite that original integral in terms of Gamma functions. Clearly to get the result we want, we want to show that I can be written in terms of the product of those two Gamma functions. But note that if we expand [itex]\Gamma(1-\epsilon)\Gamma(\epsilon)[/itex] we get a double integral whereas the expression we are given for [itex]I(c)[/itex] is just a single integral. How do I get this to work?

Thanks.
 

Answers and Replies

  • #2
fzero
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Use [itex] (x+c)^{-1} = \int_0^\infty d \alpha e^{-\alpha (x+c)} [/itex].
 
  • #3
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Use [itex] (x+c)^{-1} = \int_0^\infty d \alpha e^{-\alpha (x+c)} [/itex].

So I can substitute that into I(c) to get

[itex]I(c)=\int_0^\infty x^{1-\epsilon} \left( \int_0^\infty e^{-\alpha (x+c)} d \alpha \right) dx[/itex]

But this hasn't helped me get a [itex]\Gamma[/itex] function, unless I'm missing some blindingly obvious substitution!
 
  • #4
dextercioby
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I think that you can use Fubini's theorem and perform the x integration first.
 
  • #5
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I think that you can use Fubini's theorem and perform the x integration first.

http://en.wikipedia.org/wiki/Fubini_theorem

tells me I can write

[itex]I(c)=\int_0^\infty \left( \int_0^\infty x^{1-\epsilon} e^{-\alpha x} dx \right) e^{-\alpha c} d \alpha[/itex]

The we can use the identity [itex]\alpha^{-\lambda} \Gamma(\lambda)=\int_0^\infty dx x^{\lambda - 1} e^{-\alpha x}[/itex]
to identify [itex]1-\epsilon = \lambda-1[/itex] (comparing powers of [itex]x[/itex]
So that tells us [itex]\lambda =2-\epsilon[/itex]
and hence
[itex]I(c)=\int_0^\infty \Gamma(2-\epsilon) \alpha^{-2+\epsilon} e^{-\alpha c} d \alpha[/itex]
[itex]=\Gamma(2-\epsilon) \int_0^\infty \alpha^{-2+\epsilon} e^{-\alpha c} d \alpha [/itex]

And now we use the identity [itex]c^{-\lambda} \Gamma(\lambda) = \int_0^\infty d \alpha \alpha^{\lambda-1} e^{-\alpha c}[/itex]
to identify [itex]-2+\epsilon = \lambda-1 \Rightarrow \lambda=\epsilon-1[/itex] (comparing coefficients of [itex]\alpha[/itex]

This then gives

[itex]I(c) = \Gamma(2-\epsilon) c^{1-\epsilon} \Gamma(\epsilon-1)[/itex]

So I think (hopefully) that I have got the method right now but have just messed up the algebra?
 
  • #6
fzero
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This then gives

[itex]I(c) = \Gamma(2-\epsilon) c^{1-\epsilon} \Gamma(\epsilon-1)[/itex]

So I think (hopefully) that I have got the method right now but have just messed up the algebra?

You can use the ususal gamma function identities to put this into a form where you can apply the identity stated in the problem.
 
  • #7
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You can use the ususal gamma function identities to put this into a form where you can apply the identity stated in the problem.

Awesome - I got this to work out!

However, I then tried the more complicated example

[itex]I(c,c')=\int_0^\infty dx \frac{x^{1-\epsilon}}{(x+c)(x+c')}[/itex]
[itex]=\int)^\infty dx x^{1-\epsilon} \int_0^\infty d \alpha e^{-\alpha x} e^{-\alpha c} \int+0^\infty d \alpha' e^{-\alpha' x} e^{-\alpha' c}[/itex]
using the same relationship for [itex](x+c)^{-1}[/itex] that you gave me last time

[itex]I(c,c')= \int_0^\infty d \alpha \int_0^\infty d \alpha' \left( \int_0^\infty dx x^{1- \epsilon} e^{-x ( \alpha + \alpha')} \right) e^{-\alpha c} e^{-\alpha' c}[/itex]
[itex]= \int_0^\infty d \alpha \int_0^\infty d \alpha' (\alpha + \alpha')^{2-\epsilon} \Gamma(2-\epsilon) e^{-\alpha c} e^{-\alpha' c}[/itex]

But I need to split up the [itex]\alpha[/itex] and [itex]\alpha'[/itex] terms in [itex]( \alpha + \alpha' )^{2- \epsilon}[/itex] in order to perform those two seperate integrations.
How do I go about this?

Thanks.
 
  • #8
fzero
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You should probably simplify the original integral with partial fractions before making the substitution.
 
  • #9
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You should probably simplify the original integral with partial fractions before making the substitution.

Ok.

So I managed to get the result. It now asks me for what values of [itex]\epsilon[/itex] these are convergent - will it be [itex]\epsilon >0[/itex]? Initially I thought [itex]\epsilon \geq 0[/itex] but surely that will diverge when [itex]\epsilon=0[/itex]?

And then it says to calculate the divergent parts of both results as given by poles in [itex]\epsilon[/itex]. Well I can find the poles easily enough - these will be when [itex]\sin{\pi \epsilon}=0[/itex] i.e. [itex]\epsilon \in \pi \mathbb{Z}[/itex] but I don't really know what it means by calculate the divergent parts - what does it want me to do?

Thanks.
 
Last edited:
  • #10
fzero
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Ok.

So I managed to get the result. It now asks me for what values of [itex]\epsilon[/itex] these are convergent - will it be [itex]\epsilon >0[/itex]? Initially I thought [itex]\epsilon \geq 0[/itex] but surely that will diverge when [itex]\epsilon=0[/itex]?

You can use a comparison test. You'll have to break the integral into two parts, one that has a possible divergence at infinity and the other which possibly diverges at 0.

And then it says to calculate the divergent parts of both results as given by poles in [itex]\epsilon[/itex]. Well I can find the poles easily enough - these will be when [itex]\sin{\pi \epsilon}=0[/itex] i.e. [itex]\epsilon \in \pi \mathbb{Z}[/itex] but I don't really know what it means by calculate the divergent parts - what does it want me to do?

Thanks.

You want to compute the coefficient of [tex]1/(\epsilon-\epsilon_n)[/tex], where [tex]\epsilon_n[/tex] are the poles. You'll want to make a change of variables that lets you use [tex]\lim_{y\rightarrow 0} \sin y/y = 1[/tex] to simplify the expression that you already have in the vicinity of a pole.
 
  • #11
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You can use a comparison test. You'll have to break the integral into two parts, one that has a possible divergence at infinity and the other which possibly diverges at 0.



You want to compute the coefficient of [tex]1/(\epsilon-\epsilon_n)[/tex], where [tex]\epsilon_n[/tex] are the poles. You'll want to make a change of variables that lets you use [tex]\lim_{y\rightarrow 0} \sin y/y = 1[/tex] to simplify the expression that you already have in the vicinity of a pole.

So to sort out the convergence bit first of all, let's take the case of [itex]I(c)=\int_0^\infty \frac{x^{1-\epsilon}}{x+c}dx[/itex]

This converges iff [itex]\displaystyle\sum_{n=0}^\infty \frac{n^{1-\epsilon}}{n+c}[/itex]

So when you said split it in two, I assume you mean to consider to seperate sums

[itex]\displaystyle\sum_{n=0}^\infty \frac{1}{n+c} < \displaystyle\sum_{n=0}^\infty \frac{1}{n}[/itex] which is the bit that possible diverges at 0. However, the 1/n bit will diverge when we hit the n=0 term???

and the [itex]\displaystyle\sum_{n=0}^\infty n^{1-\epsilon}[/itex] term as the bit that possibly diverges at infinity. What should I make this less than?
I guess I could pick [itex]\epsilon >0[/itex] then i'd have it less than [itex]\displaystyle\sum_{n=0}^\infty n[/itex] but that will diverge on the [itex]n=\infty[/itex] term, no?
 
  • #12
fzero
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There's no reason to introduce sums. Write

[itex]
\int_0^\infty \frac{x^{1-\epsilon}}{x+c}dx = \int_0^1 \frac{x^{1-\epsilon}}{x+c}dx
+\int_1^\infty \frac{x^{1-\epsilon}}{x+c}dx [/itex]

Now

[tex]\frac{x^{1-\epsilon}}{x+c} < x^{-\epsilon} [/tex]

so

[tex] \int_1^\infty \frac{x^{1-\epsilon}}{x+c}dx < \int_1^\infty x^{-\epsilon} dx [/tex]

with a similar expression for the other part.
 
  • #13
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There's no reason to introduce sums. Write

[itex]
\int_0^\infty \frac{x^{1-\epsilon}}{x+c}dx = \int_0^1 \frac{x^{1-\epsilon}}{x+c}dx
+\int_1^\infty \frac{x^{1-\epsilon}}{x+c}dx [/itex]

Now

[tex]\frac{x^{1-\epsilon}}{x+c} < x^{-\epsilon} [/tex]

so

[tex] \int_1^\infty \frac{x^{1-\epsilon}}{x+c}dx < \int_1^\infty x^{-\epsilon} dx [/tex]

with a similar expression for the other part.

[itex]\frac{x^{1-\epsilon}}{x+c} < x^{1-\epsilon}[/itex] as [itex]c>0[/itex]

but surely [itex]x^{1-\epsilon} =x \cdot x^{-\epsilon} < x^{-\epsilon}[/itex] for [itex] x \in (0,1)[/itex]?
 
  • #14
fzero
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[itex]\frac{x^{1-\epsilon}}{x+c} < x^{1-\epsilon}[/itex] as [itex]c>0[/itex]

but surely [itex]x^{1-\epsilon} =x \cdot x^{-\epsilon} < x^{-\epsilon}[/itex] for [itex] x \in (0,1)[/itex]?

We're interested in the behavior as [tex]x\rightarrow \infty[/tex], so the right comparison is with

[itex]\frac{x^{1-\epsilon}}{x} =x^{-\epsilon}.[/itex]
 
  • #15
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We're interested in the behavior as [tex]x\rightarrow \infty[/tex], so the right comparison is with

[itex]\frac{x^{1-\epsilon}}{x} =x^{-\epsilon}.[/itex]

So we'd have

[itex]\int_0^1 \frac{x^{1-\epsilon}}{x+c} dx < \int_0^1 x^{-\epsilon} dx[/itex]

so we'd then want to have [itex]\epsilon > 0[/itex] in order for convergence?
 
  • #16
fzero
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So we'd have

[itex]\int_0^1 \frac{x^{1-\epsilon}}{x+c} dx < \int_0^1 x^{-\epsilon} dx[/itex]

so we'd then want to have [itex]\epsilon > 0[/itex] in order for convergence?

Does the integral converge for [tex]\epsilon=2[/tex]?
 

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