Compute the double integral I want to compute the following double integral:
$$\int_0^1dx\int_x^1xe^{y^3}dy$$
I'm can't seem to get the right answer though..
Using integration by substitution ($u = y^3$, $du = 3y^2 dy$) I get:
$$=\int_0^1dx\int_x^1\frac{x}{3y^2}e^udu=\int_0^1dx(\frac{x}{3y^2}e^{y^3})_x^1$$
I evaluate the expression from x to 1 and get:
$$=\int_0^1(\frac{ex}{3}-\frac{e^{x^3}}{3x})dx$$
I use integration by parts and for the second part I use integration by substitution ($u = x^3$, $du = 3x^2 dx$):
$$=\left[\frac{ex^2}{6}\right]_0^1-\int_0^1\frac{e^u}{9x^3}dx=\frac{e}{6}-\left[\frac{e^{x^3}}{9x^3}\right]_0^1=\frac{e}{6}-(\frac{e}{9}-\frac{1}{0})$$
And this is where it stops for me. I get a zero in one of the denominators. Where am I going wrong?
 A: The integration in $y$ is wrong: $y$ is not constant with respect to $u$. 
After the  substitution $u=y^3$, you should have
$$\int_{x=0}^1dx\int_{y=x}^1xe^{y^3}dy=\int_{x=0}^1dx\int_{u=x^{1/3}}^1xe^u\frac{du}{3y^2}=\int_{x=0}^1dx\int_{u=x^{1/3}}^1xe^u\frac{du}{3u^{2/3}}.$$
Consider that, by Fubini Theorem, here we may switch the order of integration 
$$\int_{x=0}^1dx\int_{y=x}^1 xe^{y^3}dy=\int_{y=0}^1dy\,e^{y^3} \int_{x=0}^yxdx=\frac{1}{2}\int_{y=0}^1 e^{y^3} y^2 dy.$$
Can you take it from here?
A: Well, using a more general approach:
$$\mathscr{I}_{\space\text{n}}:=\int_0^\text{n}\int_x^\text{n}x\exp\left(\text{y}^{3\text{n}}\right)\space\text{d}\text{y}\space\text{d}x=\int_0^\text{n}x\cdot\left\{\int_x^\text{n}\exp\left(\text{y}^{3\text{n}}\right)\space\text{d}\text{y}\right\}\space\text{d}x\tag1$$
Using:
$$\exp\left(\text{y}^{3\text{n}}\right):=e^{\text{y}^{3\text{n}}}=\sum_{\text{k}=0}^\infty\frac{\left(\text{y}^{3\text{n}}\right)^\text{k}}{\text{k}!}=\sum_{\text{k}=0}^\infty\frac{\text{y}^{3\text{n}\text{k}}}{\text{k}!}\tag2$$
So, we get:
$$\mathscr{I}_{\space\text{n}}=\int_0^\text{n}x\cdot\left\{\int_x^\text{n}\sum_{\text{k}=0}^\infty\frac{\text{y}^{3\text{n}\text{k}}}{\text{k}!}\space\text{d}\text{y}\right\}\space\text{d}x=\int_0^\text{n}x\cdot\left\{\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\int_x^\text{n}\text{y}^{3\text{n}\text{k}}\space\text{d}\text{y}\right\}\space\text{d}x\tag3$$
Now, we can use:
$$\int\text{a}^\text{b}\space\text{d}\text{a}=\frac{\text{a}^{1+\text{b}}}{1+\text{b}}+\text{C}\tag4$$
So, we get:
$$\mathscr{I}_{\space\text{n}}=\int_0^\text{n}x\cdot\left\{\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\left[\frac{\text{y}^{1+3\text{n}\text{k}}}{1+3\text{n}\text{k}}\right]_x^\text{n}\right\}\space\text{d}x=$$
$$\int_0^\text{n}x\cdot\left\{\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{1+3\text{n}\text{k}}\cdot\left(\text{n}^{1+3\text{n}\text{k}}-x^{1+3\text{n}\text{k}}\right)\right\}\space\text{d}x=$$
$$\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{1}{1+3\text{n}\text{k}}\cdot\int_0^\text{n}x\cdot\left(\text{n}^{1+3\text{n}\text{k}}-x^{1+3\text{n}\text{k}}\right)\space\text{d}x\tag5$$
Which gives:
$$\mathscr{I}_{\space\text{n}}=\frac{1}{6}\sum_{\text{k}=0}^\infty\frac{1}{\text{k}!}\cdot\frac{\text{n}^{3+3\text{n}\text{k}}}{1+\text{n}\text{k}}\tag6$$
When $\Re\left(1+3\text{n}\text{k}\right)>-2$
