# evaluate double integral of exponential function

I encounter some problem while evaluating this double integral. I have tried to use u-substitution and i keep using integration by parts but the equation becomes longer and longer. I found out that $$e^{-u^2}$$ is not an elementary function and if I want to integrate(indefinite integral) this, the result will be sqrt pi.(which is gaussian integral). inside the textbook I couldnt find any example similar to this. But I found examples online which most of them only dealing with one integral, this one is double integral. Could you point out on how can I evaluate this double integral? Thank you. $$\int_0^{1/2} \int_0^y e^{-(1-2x)^2} dxdy + \int_{1/2}^1\int_0^{1-y} e^{-(1-2x)^2}dxdy$$

• Have you tried changing the order of integration? Apr 23, 2017 at 16:31

We can change the order of integration (See Fubini's Theorem) to arrive at

\begin{align} \int_0^{1/2} \int_0^y e^{-(1-2x)^2} \,dx\,dy&=\int_0^{1/2} \int_x^{1/2} e^{-(1-2x)^2} \,dy\,dx\\\\ &=\int_0^{1/2}(1/2-x)e^{-(1-2x)^2}\,dx\\\\ &=\frac12\int_0^{1/2}(1-2x)e^{-(1-2x)^2}\,dx \end{align}

And similarly for the second integral

\begin{align} \int_{1/2}^1 \int_0^{1-y} e^{-(1-2x)^2} \,dx\,dy&=\int_0^{1/2} \int_{1/2}^{1-x} e^{-(1-2x)^2} \,dy\,dx\\\\ &=\int_0^{1/2}(1/2-x)e^{-(1-2x)^2}\,dx\\\\ &=\frac12\int_0^{1/2}(1-2x)e^{-(1-2x)^2}\,dx \end{align}

Can you finish?

• @cookick Well, you very welcome. And I'm pleased to hear that you have it now! -Mark ... P.S. The final answer is $\frac{1}{4}(1-e^{-1})$. Apr 23, 2017 at 17:10