Investigating the Proportional Relationship of a Gravity-Constrained Oscillator

[alejandrito29]

I has read the link

http://physics.stackexchange.com/qu...particle-under-gravity-constrained-to-move-in

and, i don't unsderstand why the integral

$$\int^1_0 \frac{dy}{\sqrt{1+y^4}}$$ is proportional to $$\Gamma (5/4) / \Gamma (3/4)$$.

I had read the definition of Beta function, but i don't get to the answer...

 

[Ray Vickson]

I has read the link

http://physics.stackexchange.com/qu...particle-under-gravity-constrained-to-move-in

and, i don't unsderstand why the integral

$$\int^1_0 \frac{dy}{\sqrt{1+y^4}}$$ is proportional to $$\Gamma (5/4) / \Gamma (3/4)$$.

I had read the definition of Beta function, but i don't get to the answer...

So, what DO you get? You are required here to show your work.

 

[alejandrito29]

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 $$\int^1_0 \frac{dy}{\sqrt{1+y^4}}$$

 

[brmath]

There is a very good hint here: http://math.stackexchange.com/questions/200739/integral-int-01-sqrt1-x4dx

 

[Dick]

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 $\int^1_0 \frac{dy}{\sqrt{1+y^4}}$.

 

[brmath]

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]

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.

 

[Ray Vickson]

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]

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