# Beta function

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Ray Vickson
Homework Helper
Dearly Missed
So, what DO you get? You are required here to show your work.
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 $$\int^1_0 \frac{dy}{\sqrt{1+y^4}}$$

Dick
Homework Helper
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 $$\int^1_0 \frac{dy}{\sqrt{1+y^4}}$$
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}}##.

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

Dick
Homework Helper
Yes, I know the hint is for minus and here we have a plus. But it seemed like the approach might be useful.
It might be, I'm just curious what problem we are actually trying to solve here. It's pretty straightforward with the minus sign.

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Ray Vickson
Homework Helper
Dearly Missed
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 $$\int^1_0 \frac{dy}{\sqrt{1+y^4}}$$
Capitalization was intended and was needed to get the meaning I wanted to convey.

lurflurf
Homework Helper
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
$$\mathrm{B}(m,n)=\int_0^\infty \frac{u^m}{(1+u)^{m+n}}\frac{\mathrm{d}u}{u}=\frac{\Gamma(n)\Gamma(m)}{\Gamma(m+n)}$$

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