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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?

Robert Z
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2 Answers2

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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?

Robert Z
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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 $\Re\left(1+3\text{n}\text{k}\right)>-2$

Jan Eerland
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