When can we merge multiple integrals
In an exercise we had to calculate $\int\limits_{-\infty}^{+\infty}\int\limits_{-\infty}^{+\infty}e^{-(x_1^2+x_2^2)}dx_1dx_2$ and deduce $\int\limits_{-\infty}^{+\infty}e^{-x^2}dx$
To do this we calculate the double integral using polar coordinates, which gives $\pi$.
The second one is calculated like this:
$\int\limits_{-\infty}^{+\infty}e^{-x^2}dx=\sqrt{\int\limits_{-\infty}^{+\infty}e^{-x_1^2}dx_1\int\limits_{-\infty}^{+\infty}e^{-x_2^2}dx_2}=\sqrt{\int\limits_{-\infty}^{+\infty}e^{-(x_1^2+x_2^2)}dx_1dx_2}=\sqrt{\pi}$
Why were we allowed to merge those two integrals?
$\endgroup$ 13 Answers
$\begingroup$Because:$$\int\int f(x)g(y)\;dy\;dx=\int f(x)\int g(y)\;dy\;dx=\int g(y)\;dy\times \int f(x)\;dx$$
Here $f(x)=g(x)=e^{-x^2}$.
$\endgroup$ 3 $\begingroup$note that what you had was:$$\int_a^b\int_a^bf(x)f(y)dxdy$$note that each function is independent and so we get:$$\left[F(x)\right]_a^b\left[F(y)\right]_a^b=\left(\left[F(z)\right]_a^b\right)^2$$
$\endgroup$ $\begingroup$If the limits of integration are independent of the variables...
$\int_a^b\int_c^d f(x)g(y)\ dy\ dx = (\int_a^b f(x)\ dx)(\int_c^d g(y)\ dy)$
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