# Definite integral and Gamma functions

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## Main Question or Discussion Point

I've been trying to determine how certain definite integrals are expressed in terms of Gamma functions.

Mathematica returns the following:

$$\int_0^1 \frac{dx}{\sqrt{1-x^4}}=\frac{\sqrt{\pi}\Gamma[\frac{5}{4}]}{\Gamma[\frac{3}{4}]}$$

(Mapple returns a different but equivalent expression in terms of Gamma)

In general it seems:

$$\int_0^1 \frac{dx}{\sqrt{1-x^n}}=\frac{\sqrt{\pi}\Gamma[\frac{n+1}{n}]}{\Gamma[\frac{n+2}{2n}]}$$

Can anyone explain to me how this is determined or provide a hint or a reference?

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dextercioby
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According to Mathematica,the general case is a standard integral for the Gauss hypergeometric function...

Daniel.

Well, the trick is to convert the given integral into gamma-integrals. If i remember correctly from my calculus course at college, you can do this via PI...though i am not too sure

marlon

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marlon said:
Well, the trick is to convert the given integral into gamma-integrals. If i remember correctly from my calculus course at college, you can do this via PI...though i am not too sure

marlon
Well thanks Marlon and Daniel, but via pi?

You mean, I need to figure out how to convert:

$$\int_0^1\frac{dx}{\sqrt{1-x^4}}$$

to some variant of:

$$\Gamma[x]=\int_0^{\infty}e^{-t}t^{x-1}dt$$

Well, I'll look in my Calculus books but well, I just don't see it happening. Maybe so though. Think you can give me another hint? It's a very interesting problem and I'd like to know how to figure it out.

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dextercioby
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I don't see how u could possibly put an exponential there.I'm waiting anxiously to be stunned...

Daniel.

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Here we go (on the web under "gamma integral"):

$$I=\int_0^1\frac{dx}{(1-x^a)^b}$$

with a>0 and b<1

Letting:

$$u=x^a$$

then:

$$dx=\frac{1}{a}u^{(1/a)-1}$$

So that we now have:

$$\frac{1}{a}\int_0^1 u^{(1/a)-1}(1-u)^{-b}du$$

Now, here's the key: The Beta function is defined:

$$\beta(m,n)=\int_0^1 u^{m-1}(1-u)^{n-1}du$$

So that the integral, expressed in the beta function is:

$$I=\frac{1}{a}\beta(\frac{1}{a},1-b)$$

Since:

$$\beta(m,n)=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$$

We finally have:

$$I=\frac{\Gamma(\frac{1}{a})\Gamma(1-b)}{a\Gamma(1-b+\frac{1}{a})}$$

Think I need to reiew this with a couple of examples . . .

Edit: Also, made an error with the upper limit in the original post, it should have been from 0 to 1.

Last edited:
dextercioby