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.
$$\int_0^{1/2} \int_0^y e^{-(1-2x)^2} dxdy + \int_{1/2}^1\int_0^{1-y} e^{-(1-2x)^2}dxdy $$
[Math] evaluate double integral of exponential function
integration
Best Answer
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?