# Thread: How to transform double integral into a single integral using diamond transform

1. ## How to transform double integral into a single integral using diamond transform

I have a double integral given as,
$\frac{1}{T^{2}}\int_{-\frac{T}{2}}^{\frac{T}{2}}\int_{-\frac{T}{2}}^{\frac{T}{2}}\rho(t_{2}-t_{1})dt_{1}dt_{2}$ and I would like to convert it into a single integral as $\frac{2}{T}\int_{0}^{T}\left(1-\frac{\tau}{T}\right)\rho(\tau)d\tau$, where $\tau = t_{2}-t_{1}$. I read in the book that diamond transform is used to convert the given double integral into the single integral but am unable to understand it, since no elaboration is given. Can someone help me to understand how this double integral is transformed to the single integral as given above?

Thanks,
Ameya

2. Enough with the taus and stuff and should $\rho(x)$ not be an even function for that? Suppose so then how about just x, and y and u and v and just a real example with T=2 which in that case:

$\frac{1}{4}\int_{-1}^{1} \int_{-1}^{1} f(y-x)dx dy$

which is a square. Now if I let $u=y-x$ and $v=y+x$ then that transforms the region in x and y to a diamond shape in the u-v plane going from (2,0) to (0,2) then (-2,0) to (0,-2) and back to (2,0). Then calculate the Jacobian $\frac{\partial (x,y)}{\partial (u,v)}=-1/2$ but the positive direction over the u-v contour goes in the negative direction over the x-y contour so drop the negative sign. Then I get:

$\frac{2}{2}\int_0^{2} (1-\frac{u}{2}) f(u)du$

assuming $f(x)$ is an even function and is in the form you stated. Looks to me anyway.

3. ## Question

It's been awhile since I took calculus and this is the first time I'm hearing about a diamond transform. Can someone explain what this is?

4. Originally Posted by shawsend
Enough with the taus and stuff and should $\rho(x)$ not be an even function for that? Suppose so then how about just x, and y and u and v and just a real example with T=2 which in that case:

$\frac{1}{4}\int_{-1}^{1} \int_{-1}^{1} f(y-x)dx dy$

which is a square. Now if I let $u=y-x$ and $v=y+x$ then that transforms the region in x and y to a diamond shape in the u-v plane going from (2,0) to (0,2) then (-2,0) to (0,-2) and back to (2,0). Then calculate the Jacobian $\frac{\partial (x,y)}{\partial (u,v)}=-1/2$ but the positive direction over the u-v contour goes in the negative direction over the x-y contour so drop the negative sign. Then I get:

$\frac{2}{2}\int_0^{2} (1-\frac{u}{2}) f(u)du$

assuming $f(x)$ is an even function and is in the form you stated. Looks to me anyway.
Thank you very much for your reply. This is indeed what I was looking for. You are right; $f(x)$ is indeed an even function.

5. ## Re: How to transform double integral into a single integral using diamond transform

Can someone explain how is 1-u/2 found?

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