# Gamma Functions

Consider $I(c)= \int_0^\infty \frac{x^{1-\epsilon}}{x+c} dx$

For what values of $\epsilon$ are these integrals convergent?

Would it be $\epsilon \geq 0$?

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

$I(c)=-\frac{\pi}{\sin{\pi \epsilon}}c^{1-\epsilon}$

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 $\Gamma(1-\epsilon)\Gamma(\epsilon)$ we get a double integral whereas the expression we are given for $I(c)$ is just a single integral. How do I get this to work?

Thanks.

## Answers and Replies

fzero
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Use $(x+c)^{-1} = \int_0^\infty d \alpha e^{-\alpha (x+c)}$.

Use $(x+c)^{-1} = \int_0^\infty d \alpha e^{-\alpha (x+c)}$.

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

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

But this hasn't helped me get a $\Gamma$ function, unless I'm missing some blindingly obvious substitution!

dextercioby
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I think that you can use Fubini's theorem and perform the x integration first.

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

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

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

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

This then gives

$I(c) = \Gamma(2-\epsilon) c^{1-\epsilon} \Gamma(\epsilon-1)$

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

fzero
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This then gives

$I(c) = \Gamma(2-\epsilon) c^{1-\epsilon} \Gamma(\epsilon-1)$

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.

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

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

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

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

Thanks.

fzero
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You should probably simplify the original integral with partial fractions before making the substitution.

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 $\epsilon$ these are convergent - will it be $\epsilon >0$? Initially I thought $\epsilon \geq 0$ but surely that will diverge when $\epsilon=0$?

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

Thanks.

Last edited:
fzero
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Ok.

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

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 $\epsilon$. Well I can find the poles easily enough - these will be when $\sin{\pi \epsilon}=0$ i.e. $\epsilon \in \pi \mathbb{Z}$ 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 $$1/(\epsilon-\epsilon_n)$$, where $$\epsilon_n$$ are the poles. You'll want to make a change of variables that lets you use $$\lim_{y\rightarrow 0} \sin y/y = 1$$ to simplify the expression that you already have in the vicinity of a pole.

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 $$1/(\epsilon-\epsilon_n)$$, where $$\epsilon_n$$ are the poles. You'll want to make a change of variables that lets you use $$\lim_{y\rightarrow 0} \sin y/y = 1$$ 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 $I(c)=\int_0^\infty \frac{x^{1-\epsilon}}{x+c}dx$

This converges iff $\displaystyle\sum_{n=0}^\infty \frac{n^{1-\epsilon}}{n+c}$

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

$\displaystyle\sum_{n=0}^\infty \frac{1}{n+c} < \displaystyle\sum_{n=0}^\infty \frac{1}{n}$ 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 $\displaystyle\sum_{n=0}^\infty n^{1-\epsilon}$ term as the bit that possibly diverges at infinity. What should I make this less than?
I guess I could pick $\epsilon >0$ then i'd have it less than $\displaystyle\sum_{n=0}^\infty n$ but that will diverge on the $n=\infty$ term, no?

fzero
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There's no reason to introduce sums. Write

$\int_0^\infty \frac{x^{1-\epsilon}}{x+c}dx = \int_0^1 \frac{x^{1-\epsilon}}{x+c}dx +\int_1^\infty \frac{x^{1-\epsilon}}{x+c}dx$

Now

$$\frac{x^{1-\epsilon}}{x+c} < x^{-\epsilon}$$

so

$$\int_1^\infty \frac{x^{1-\epsilon}}{x+c}dx < \int_1^\infty x^{-\epsilon} dx$$

with a similar expression for the other part.

There's no reason to introduce sums. Write

$\int_0^\infty \frac{x^{1-\epsilon}}{x+c}dx = \int_0^1 \frac{x^{1-\epsilon}}{x+c}dx +\int_1^\infty \frac{x^{1-\epsilon}}{x+c}dx$

Now

$$\frac{x^{1-\epsilon}}{x+c} < x^{-\epsilon}$$

so

$$\int_1^\infty \frac{x^{1-\epsilon}}{x+c}dx < \int_1^\infty x^{-\epsilon} dx$$

with a similar expression for the other part.

$\frac{x^{1-\epsilon}}{x+c} < x^{1-\epsilon}$ as $c>0$

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

fzero
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$\frac{x^{1-\epsilon}}{x+c} < x^{1-\epsilon}$ as $c>0$

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

We're interested in the behavior as $$x\rightarrow \infty$$, so the right comparison is with

$\frac{x^{1-\epsilon}}{x} =x^{-\epsilon}.$

We're interested in the behavior as $$x\rightarrow \infty$$, so the right comparison is with

$\frac{x^{1-\epsilon}}{x} =x^{-\epsilon}.$

So we'd have

$\int_0^1 \frac{x^{1-\epsilon}}{x+c} dx < \int_0^1 x^{-\epsilon} dx$

so we'd then want to have $\epsilon > 0$ in order for convergence?

fzero
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So we'd have

$\int_0^1 \frac{x^{1-\epsilon}}{x+c} dx < \int_0^1 x^{-\epsilon} dx$

so we'd then want to have $\epsilon > 0$ in order for convergence?

Does the integral converge for $$\epsilon=2$$?