Finding the Product of Integrals

[drewfstr314]

Is there a formula for calculating the product of integrals, something like:

\left(\int_a^b f(x) dx\right) \times \left(\int_c^d g(y) dy\right)

when there is no closed-form expression for F(x) and G(y).

Actually, the functions are almost identical,

f(x) = x^p e^{-x} \text{ and } g(y) = y^q e^{-y}

where p, q are algebraic expressions.

F(x) = -\Gamma(x, p) \text{ and } G(x) = -\Gamma(y, q)

and \Gamma(x, p) is defined as another definite integral with an almost identical integrand.

Thus, is there a way to multiply definite integrals (without knowing the antiderivative) to form one (double?) integral?

Thanks

 

[Millennial]

If the limits are equal, yes. Then it is converted as follows:
\int^{a}_{b}f(x)dx \cdot \int^{a}_{b}g(y)dy = \int^{a}_{b}\int^{a}_{b}f(x)g(y)\,dy\,dx

 

[drewfstr314]

If the limits are equal, yes. Then it is converted as follows:
\int^{a}_{b}f(x)dx \cdot \int^{a}_{b}g(y)dy = \int^{a}_{b}\int^{a}_{b}f(x)g(y)\,dy\,dx

Is the new integral a double integral then?

 

[Millennial]

Yes it is.

 

[haruspex]

If the limits are equal, yes. Then it is converted as follows:
\int^{a}_{b}f(x)dx \cdot \int^{a}_{b}g(y)dy = \int^{a}_{b}\int^{a}_{b}f(x)g(y)\,dy\,dx

There's no requirement that the limits match. As long as the variables and ranges involved are completely independent, you can always combine them into a double integral (in either order). Just make sure you keep track of which independent variable goes with which range.

 

[HallsofIvy]

That's "Fubini's theorem"