Double integral $\int_0^1 \int_{\sqrt{x}}^1 \sin(\pi y^3)\,dy\,dx$

Question

Calculate the integral: $$\int_0^1(\int_{\sqrt{x}}^1\sin(\pi y^3)dy)dx$$

I have come up with this solution:

If $\sqrt{x} \leq y \leq 1$ then $x \leq y^2 \leq 1$ and since $0 \leq x \leq 1$ we have that $0 \leq y \leq 1$ and $0 \leq x \leq y^2$.

If we insert this into the integral we get: $$\int_0^{y^2}(\int_{0}^1\sin(\pi y^3)dy)dx = \int_{0}^1(\int_0^{y^2}sin(\pi y^3)dx)dy$$

This I can easily calculate using variable substitution ($u = y^3; du = 2y^2dx $ giving me the result $\frac{2}{3\pi}$, I dont have the key).

However I'm not sure what I'm doing with my integration limits is "legal", so to say.

There is also a follow up question:

"Is the following true or false? Motivate your answer (without long calculations)."

$$\int_0^1(\int_{\sqrt{x}}^1\sin(\pi y^4)dy)dx = 0$$

Just looking at it my answer would be false, but I have no good reason for that.

I feel like there is some theory I'm missing here.

Thanks in advance and sorry for my poor english.

Answer 1

To the first question: Indeed, the first integral equals $\frac23\pi$. In fact, you are applying Tonelli's Theorem while integrating over the set $\{(x,y)\in[0,1]\times[0,1]\mid x\le y^2 \}$.

To the second question: The integrand doesn't vanish, since $\sin(\pi y^4)>0$ for $0<y<1$, so by monotonicity of the integral $$I(x):=\int_{\sqrt{x}}^1 \sin(\pi y^4)\,\mathrm dy>0$$ for $0\le x<1$ and thus by monotonicity again: $$\int_0^1 I(x)\,\mathrm dx > 0.$$

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The integral can't be expressed in terms of elementary functions, numerical methods give

$$\int_0^1 I(x)\,\mathrm dx\approx0.197788.$$

Answer 2

Note that, your integrand is non-negative and hence by Tonelli's theorem switching the order of integration is justified. Same non-negativity (the integrand is actually positive on a domain of positive measure inside the region of integration) tells you that the claim can't be true.