Computing and proving existence of integral We're supposed to prove the existence of the following integral and compute it.
$\int^{10}_{y=0}\int^{\pi/3}_{x=0}{xy \cos x y^2}\,\mathrm{d}x\mathrm{d}y$
My two cents on this. I was trying to use Fubini's theorem and change the order of integration, but ran into trouble evaluating the actual integral itself.
 A: It exist because the integrand is continuous on the compact set $[0,\pi/3]\times [0,10]$. Then you can apply Fubini's Theorem and get
\begin{align}\int^{10}_{y=0}\int^{\pi/3}_{x=0}{xy \cos x y^2}\,\mathrm{d}x\mathrm{d}y&=\int^{\pi/3}_{x=0}\left(\int^{10}_{y=0}{xy \cos x y^2}\,\mathrm{d}y\right)\mathrm{d}x\\
&=\int^{\pi/3}_{x=0}\left[\frac{1}{2} \sin (x y^2)\right]^{10}_{y=0}\mathrm{d}x.
\end{align}
Can you take it from here?
A: Well, we know that:
$$\mathscr{I}:=\int_0^{10}\int_0^\frac{\pi}{3}x\cdot\text{y}\cdot\cos\left(x\cdot\text{y}^2\right)\space\text{d}x\space\text{d}\text{y}=$$
$$\int_0^{10}\text{y}\cdot\left\{\int_0^\frac{\pi}{3}x\cdot\cos\left(x\cdot\text{y}^2\right)\space\text{d}x\right\}\space\text{d}\text{y}\tag1$$
Using:
$$\cos\left(x\cdot\text{y}^2\right)=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\left(x\cdot\text{y}^2\right)^{2\text{n}}=$$
$$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot x^{2\text{n}}\cdot\text{y}^{4\text{n}}\tag2$$
So, we can write:
$$\mathscr{I}=\int_0^{10}\text{y}\cdot\left\{\int_0^\frac{\pi}{3}x\cdot\left(\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot x^{2\text{n}}\cdot\text{y}^{4\text{n}}\right)\space\text{d}x\right\}\space\text{d}\text{y}=$$
$$\mathscr{I}=\int_0^{10}\text{y}\cdot\left\{\int_0^\frac{\pi}{3}\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot x^{1+2\text{n}}\cdot\text{y}^{4\text{n}}\space\text{d}x\right\}\space\text{d}\text{y}=$$
$$\mathscr{I}=\int_0^{10}\text{y}\cdot\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\text{y}^{4\text{n}}\cdot\left\{\int_0^\frac{\pi}{3}x^{1+2\text{n}}\space\text{d}x\right\}\space\text{d}\text{y}\tag3$$
Now, using:
$$\int_0^\frac{\pi}{3}x^{1+2\text{n}}\space\text{d}x=\left[\frac{x^{2+2\text{n}}}{2+2\text{n}}\right]_0^\frac{\pi}{3}=\frac{\left(\frac{\pi}{3}\right)^{2+2\text{n}}}{2+2\text{n}}\tag4$$
So:
$$\mathscr{I}=\int_0^{10}\text{y}\cdot\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\text{y}^{4\text{n}}\cdot\left\{\frac{\left(\frac{\pi}{3}\right)^{2+2\text{n}}}{2+2\text{n}}\right\}\space\text{d}\text{y}=$$
$$\int_0^{10}\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\text{y}^{1+4\text{n}}\cdot\left\{\frac{\left(\frac{\pi}{3}\right)^{2+2\text{n}}}{2+2\text{n}}\right\}\space\text{d}\text{y}=$$
$$\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\frac{\left(\frac{\pi}{3}\right)^{2+2\text{n}}}{2+2\text{n}}\cdot\int_0^{10}\text{y}^{1+4\text{n}}\space\text{d}\text{y}\tag5$$
Now, using:
$$\int_0^{10}\text{y}^{1+4\text{n}}\space\text{d}\text{y}=\left[\frac{\text{y}^{2+4\text{n}}}{2+4\text{n}}\right]_0^{10}=\frac{10^{2+4\text{n}}}{2+4\text{n}}\tag6$$
So:
$$\mathscr{I}=\sum_{\text{n}=0}^\infty\frac{\left(-1\right)^\text{n}}{\left(2\text{n}\right)!}\cdot\frac{\left(\frac{\pi}{3}\right)^{2+2\text{n}}}{2+2\text{n}}\cdot\frac{10^{2+4\text{n}}}{2+4\text{n}}=\frac{3}{400}\tag7$$
