I am trying to integrate a function -
$\int_0^1 \int_y^1 {(x^r+y^r) }^{1/r}\,dx\,dy$
Here, r lies in [0,1]
I tried to solve it using substitution but to no avail. How can I solve it?
Here's what I have tried so far -
Taken u = $x^r+y^r$ and v=y
Solving for x gives -
x= ${(u-v^r)}^{1/r}$
Jacobian function for the above is - ${{(1/r)}(u-v^r)}^{{1/r}-1}$
So, the Integration function becomes -
${{(1/r)}(u-v^r)}^{{1/r}-1}{u^{1/r}}$ which is not any easier to solve.
I have tried various substitutions like
u=$x^r+y^r$ ; v=$x^r-y^r$
u=$x/y$; v= $xy$
I can share all these results too but they are generating very complex equations as well.
I don't want the solution. If anyone can point out my mistake and give some hint on how to proceed, I'll be deeply grateful.
PS: This is not a homework question.
Also, I used desmos to see whether solution exist by plugging random values of r and it existed for all r in (0,1].
Please direct me.
This is not the most pleasant integral except if you are familiar with hypergeomatric functions.
First, $$\int\left(x^r+y^r\right)^{\frac{1}{r}}\,dx=x y \, _2F_1\left(-\frac{1}{r},\frac{1}{r};\frac{1+r}{r};-\left(\frac{x}{y}\right)^r\right)$$ where appears the gaussian hypergeometric function. So, the inner definite integral is $$I(y)=\int_y^1\left(x^r+y^r\right)^{\frac{1}{r}}\,dx=$$$$y \, _2F_1\left(-\frac{1}{r},\frac{1}{r};\frac{1+r}{r};-y^{-r}\right)-y^2 \, _2F_1\left(-\frac{1}{r},\frac{1}{r};\frac{1+r}{r};-1\right)$$ Now, start complicated stuff; a CAS gives the antiderivative (too long to be printed here) but the definite integral is $$\frac{1}{6} \left(\, _2F_1\left(-\frac{2}{r},-\frac{1}{r};\frac{r-2}{r};-1\right)-\frac{\Gamma \left(\frac{1}{r}\right) \Gamma \left(\frac{r-2}{r}\right)}{\Gamma \left(-\frac{1}{r}\right)}\right)$$