# 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.

fzero
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Gold Member
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
Homework Helper
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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Gold Member
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.

Thanks.

fzero
Homework Helper
Gold Member
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.

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fzero
Homework Helper
Gold Member
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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Gold Member
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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Gold Member
$\frac{x^{1-\epsilon}}{x+c} < x^{1-\epsilon}$ 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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Gold Member
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 $$\epsilon=2$$?