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Whenever I integrate a gaussian function, I get to a step that makes me a little uncomfortable because I don't fully understand it. The only way I know of to analytically integrate the gaussian function is to multiply two of them together, like so...

$$\int_{- \infty}^{\infty} e^{-x^2}dx\int_{- \infty}^{\infty} e^{-y^2}dy = \int_{- \infty}^{\infty}\int_{- \infty}^{\infty}e^{-x^2-y^2}\,dx\,dy$$

My questions is, what allows me to group two different integrands together into one integrand? When is this technique not allowed? Thanks in advance.

Michael Hardy
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Cactus BAMF
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  • In general, if you have a separable integrand, such as $\iiint f(x)g(y)h(z),\mathrm dx\mathrm dy\mathrm dz$, and the multiple integral is sensible, you can always perform a factorization into $\left(\int f(x),\mathrm dx\right)\left(\int g(y),\mathrm dy\right)\left(\int h(z),\mathrm dz\right)$ – J. M. ain't a mathematician May 30 '13 at 02:57

4 Answers4

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You are allowed to do this due to Fubini's theorem

Heberto del Rio
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You can use this technique whenever the integrals converge.

response
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Two things are special here. Aside from the worry about improper integrals, you can express the double integral $$\iint_R f(x,y)\, dA$$ as a product of single integrals when (1) $R$ is a rectangle, and (2) $f(x,y)=g(x)h(y)$ for some functions $g$ and $h$. To spare ourselves anomalies, let's assume all functions in question are continuous.

Ted Shifrin
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You have $$ \int_a^b \left( \int_c^d f(x)g(y)\,dx\right)\,dy. $$ Look at the inside integral: $$ \int_c^d f(x)g(y)\,dx. $$ The thing you're integrating is $g(y)$ times something, and you have $x$ going from $c$ to $d$. As $x$ goes from $c$ to $d$, notice that $g(y)$ DOES NOT CHANGE. $g(y)$ does not depend on $x$. And $g(y)$ is a FACTOR of the function you're integrating, i.e. what you're integrating is $g(y)$ times something. Since this factor does not depend on $x$, it's a CONSTANT as a function of $x$, so it can be pulled out, getting $$ g(y)\int_c^d f(x)\,dx. $$ Now you have $$ \int_a^b g(y)\left(\int_c^d f(x)\,dx\right) \,dy. $$ Now notice that the expression in parentheses does not depend on $y$, and you're integrating with respect to $y$ as $y$ goes from $a$ to $b$. In other words, as a function of $y$, that expression is a CONSTANT, so it can likewise be pulled out. Then you're done.

Michael Hardy
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