Could someone help me with this question? I am stuck on it.

Compute the following double integral: $$\int_{y=0}^1\int_{x=y}^1 e^{\large x^2}\ dx\ dy.$$

How to compute the integral when the inner integral is a form of error function? Thanks in advance for your help.


The region of integration $y<x<1$ and $0<y<1$ is corresponding to $0<y<x$ and $0<x<1$, therefore $$ \int_{y=0}^1\int_{x=y}^1\ e^{\large x^2}\ dx\ dy=\int_{x=0}^1\int_{y=0}^x\ e^{\large x^2}\ dy\ dx=\int_{x=0}^1\ e^{\large x^2}\ dx\ \int_{y=0}^x\ \ dy. $$ The last integral is easy to be dealth with and the answer is $\color{blue}{\dfrac12(e-1)}$.

  • $\begingroup$ Thank you very much for your answer. One more question, is it possible to compute the inner integral without changing the limit of integral? $\endgroup$ – user151804 May 18 '14 at 15:19
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    $\begingroup$ @user151804 Please take a look my another answer. Anyway, I really enjoy your problem specially for obtaining the another answer. :) $\endgroup$ – Tunk-Fey May 18 '14 at 16:18

Another hard way (As the OP's wish):

Using Maclaurin series of exponential function, we will obtain $$ e^{\large x^2}=\sum_{n=0}^\infty\frac{x^{\large 2n}}{n!}. $$ Hence $$ \begin{align} \int_{x=y}^1\ e^{\large x^2}\ dx&=\int_{x=y}^1\ \sum_{n=0}^\infty\frac{x^{\large 2n}}{n!}\ dx\\ &=\sum_{n=0}^\infty\int_{x=y}^1\ \frac{x^{\large 2n}}{n!}\ dx\\ &=\left.\sum_{n=0}^\infty\frac{x^{\large 2n+1}}{(2n+1)\ n!}\right|_{x=y}^1\\ &=\sum_{n=0}^\infty\frac{1-y^{\large 2n+1}}{(2n+1)\ n!} \end{align} $$ and $$ \begin{align} \require{cancel} \int_{y=0}^1\ \sum_{n=0}^\infty\frac{1-y^{\large 2n+1}}{(2n+1)\ n!}\ dy&=\sum_{n=0}^\infty\int_{y=0}^1\ \frac{1-y^{\large 2n+1}}{(2n+1)\ n!}\ dy\\ &=\sum_{n=0}^\infty\left.\left(\frac{y}{(2n+1)\ n!}-\frac{y^{\large 2n+2}}{(2n+2)(2n+1)\ n!}\right)\right|_{y=0}^1\\ &=\sum_{n=0}^\infty\left(\frac{1}{(2n+1)\ n!}-\frac{1}{(2n+2)(2n+1)\ n!}\right)\\ &=\sum_{n=0}^\infty\frac{(2n+2)-1}{(2n+2)(2n+1)\ n!}\\ &=\sum_{n=0}^\infty\frac{\cancel{2n+1}}{2(n+1)\cancel{(2n+1)}\ n!}\\ &=\frac12\sum_{n=0}^\infty\frac{1}{(n+1)!}\\ &=\large\color{blue}{\frac12(e-1)}. \end{align} $$

  • $\begingroup$ @user151804 You're welcome. :) $\endgroup$ – Tunk-Fey May 18 '14 at 16:30

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