# Beta function

1. Dec 28, 2013

### alejandrito29

2. Dec 28, 2013

### Ray Vickson

3. Dec 28, 2013

### alejandrito29

ok, is not necessary write with capital letter the word "DO".

For to solve the integral $$\int^1_0 \frac{dy}{\sqrt{1+y^4}}$$ i tried to use the beta function $$\beta (x,y) = \int^1_0 t^{x-1}(1-t)^{y-1} dt$$

but, i don't find values of x and y but for to get of the form $$\int^1_0 \frac{dy}{\sqrt{1+y^4}}$$

4. Dec 28, 2013

### brmath

5. Dec 28, 2013

### Dick

That stackexchange thread is talking about the integral $\int^1_0 \frac{dy}{\sqrt{1-y^4}}$ not $\int^1_0 \frac{dy}{\sqrt{1+y^4}}$. If that's the one you really want to do then substituting $t=y^4$ is a good first step. Beta functions won't help with $\int^1_0 \frac{dy}{\sqrt{1+y^4}}$.

Last edited: Dec 28, 2013
6. Dec 28, 2013

### brmath

Yes, I know the hint is for minus and here we have a plus. But it seemed like the approach might be useful.

7. Dec 28, 2013

### Dick

It might be, I'm just curious what problem we are actually trying to solve here. It's pretty straightforward with the minus sign.

Last edited: Dec 28, 2013
8. Dec 28, 2013

### Ray Vickson

Capitalization was intended and was needed to get the meaning I wanted to convey.

9. Dec 29, 2013

### lurflurf

There are many ways to transform this, for example

$$\int_0^1 \frac{\mathrm{d}y}{\sqrt{1+y^4}}=\frac{1}{8}\int_0^\infty \frac{u^{1/4}}{\sqrt{1+u}}\frac{\mathrm{d}u}{u}=\frac{3}{2}\int_0^\infty \frac{u^{5/4}}{(1+u)^{5/2}}\frac{\mathrm{d}u}{u}$$
recall that
$$\mathrm{B}(m,n)=\int_0^\infty \frac{u^m}{(1+u)^{m+n}}\frac{\mathrm{d}u}{u}=\frac{\Gamma(n)\Gamma(m)}{\Gamma(m+n)}$$