Using double integrals to evaluate single integrals

autodidude

In most calculus textbooks, they use double integrals to evaluate the Gaussian integral. Where did they get the idea - or how did they choose the two variable function [tex]e^{-(x^2+y^x)}[/tex] to evaluate it?

I guess this is related...but if you were given a fairly hairy integral and it was suggested that you write the integrand as an integral, how would you go about doing so? The example I have in mind is [tex]\int^2_0 arctan(\pi x)-arctan(x) dx[/tex]. The only way I know how to do that is by integration by parts

mfb

Intuition, or just brute force: Test options until one of them is successful.

arctan has a nice representation as integral.

HallsofIvy

If you have [itex]\int_a^b F(x)dx[/itex] and F'(x)= f(x) then, by the fundamental theorem of Calculus,[itex]F(x)= \int_p^x f(y)dy[/tex] for some p so [itex]\int_a^b F(x)dx= \int_a^b\int_p^x f(y)dydx[/itex]

autodidude

This wouldn't be known as 'integration under the integral sign' would it? I just came across the technique in Woods' 'Advanced Calculus' and it looks similar.

I tried it on the integral I posted and I've probably misunderstood something, but it doesn't seem to simplify it at all (still have to use integration by parts). And secondly, how do you choose the lower bound p?

What I got was:

[tex]\int_0^{\pi} \int_0^2 arctan(\alpha x)-arctan(x)dxd\alpha[/tex]

I ended up choosing 0 as a lower bound cause that's when the integrand is 0.....hmmm...I think I've got down the wrong track here

mathman

I presume you realize you have a typo. The exponent is x²+y². The integration is easily carried out by converting to polar coordinates. The exponent is now r², the differential is rdrdθ, and the limits are [0,∞) and [0,2π). As you can see the integration is straightforward.

Your other example seems to be completely unrelated.

Mute

You're not quite going in the right direction. You want to start with

$$\int_0^2 dx~(\mbox{arctan}(\pi x) - \mbox{arctan}(x) ) = \int_0^2 dx~\left. \mbox{arctan}(yx)\right|^{y=\pi}_{y=1}.$$

You want to replace the ##\left. \mbox{arctan}(yx)\right|^{y=\pi}_{y=1}## with an integral. Do you think you know what to do now?

Robert1986

No, I don't think his other example is unrelated at all. Why do you think it is?

There are integrals that are difficult to evaluate, but sometimes the integrand can be modified by adding another variable. Then, one can use, say, Fubini's Theorem to evaluate the original integral by switching order of integration and comparing. So, perhaps the integral he posted can be computed that way.

Jim Kata

well the single integral is just area under the curve

so if [tex]A=\int e^{-x^2}dx[/tex]

then [tex]A^2=(\int e^{-x^2}dx)(\int e^{-y^2}dy)=\iint e^{-(x^2+y^2)}dxdy[/tex]

so[tex] A= \sqrt{\iint e^{-(x^2+y^2)}dxdy}[/tex]

autodidude

Sorry for the late reply, I've been a bit occupied with coursework.

I'm not sure how you knew to replace the original integrand with arctan(yx) evaluated from y=1 to y=π

Mute

Well, you want to make the single integral into a double integral, so it seems like the most straightforward way to do that would be to replace the integrand with another integral. You integrand had the form f(bx) - f(ax). This looks like the result of a definite integration:

$$ \int_a^b dy~\frac{d}{dy} f(yx)= \left.f(yx)\right|_{y=a}^{y=b} = f(bx) - f(ax).$$

I wrote it as ##\left. \mbox{arctan}(yx)\right|_{y=1}^{y=\pi}## rather than as an integral so that you could try and fill in the last step yourself.

Does this make more sense?