Understanding the Computation of Double Integrals: Can You Help?

[inviziblesoul]

Can anybody please help me understand the computation of the integral in the attached image. I shall be grateful.

 

[HallsofIvy]

What exactly is the integral you are asking about? The integral on the left is simple, just the integral of c(t_1- t_2) over the square -T\le t_1\le T and -T\le t_2\le T. Is c a constant or a function of t_1- t_2? If a function, then what function?

Perhaps you are trying to convert the function c(t_1- t_2) to a function of a single variable by defining \tau= t_1- t_2? The region is still two dimensional and even though c can be set as a function of one variable, you will need another variable, say \sigma= t_1+ t_2. Then t_1= (\tau+ \sigma)/2 and t_2= (\sigma- \tau)/2 so that the boundaries of the square become t_1= (\tau+ \sigma)/2= T so \tau+ \sigma= 2T, t_1= (\tau+ \sigma)/2= -T so \tau+ \sigma= -2T, t_2= (-\tau+ \sigma)/2= T so -\tau+ \sigma= 2T, and t_1= (-\tau+ \sigma)/2= -T so \tau- \sigma= 2T.

That is a rectangle in the \tau, \sigma plane with its digonals parallel to the axes. Overall, \tau goes from -2T to 2T but because the upper and lower boundaries change formula at \tau= 0 you should do it as two separate integrals. For \tau= -2T to \tau= 0, the lower boundary is \tau+ \sigma= 2T or \sigma= 2T- \tau and the upper boundary is -\sigma+ \tau= 2T or \sigma= \tau- 2T. For \tau= 0 to \tau= 2T, the lower boundary is =-\sigma+ \tau= -2T or \sigma= -\tau- 2T and the upper boundary is sigma+ \tau= 2T or \sigma= -\tau+ 2T.

So the integral can be written
\int_{-2T}^0\int_{2T-\tau}^{\tau+ 2T} c(\tau)d\sigma d\tau+ \int_0^{2T}\int_{-\tau- 2T}^{\tau- 2T} c(\tau) d\sigma d\tau

 

[inviziblesoul]

Thank you very much for your excellent efforts and this great explanation. However, I am not clear at certain points.

You have rightly pointed out: the aim here is to express the double integral in terms of a single integral. Furthermore, C is a function of the difference \tau=t_1−t_2.

I did not understand your phrase <<That is a rectangle in the τ, σ plane with its digonals parallel to the axes.>> How do you know that its a rectangle and its diagonals are parallel to the axes (the \tau, \sigma axes ?).
and how did you choose \sigma = t_1 + t_2? why not some other function?
I will greatly appreciate if you can kindly refer me some reading on this topic.

I have attached my solution as well. I have not introduced a new variable, however, I have used \tau and t_1.

Thank you for your time.

 

[HallsofIvy]

Thank you very much for your excellent efforts and this great explanation. However, I am not clear at certain points.

You have rightly pointed out: the aim here is to express the double integral in terms of a single integral. Furthermore, C is a function of the difference \tau=t_1−t_2.

I did not understand your phrase <<That is a rectangle in the τ, σ plane with its digonals parallel to the axes.>> How do you know that its a rectangle and its diagonals are parallel to the axes (the \tau, \sigma axes ?).

The "coordinate axes" in a \tau, \sigma coordinate system are the lines \tau= 0 and \sigma= 0 which mean t_1- t_2= 0 and lines parallel to that.

and how did you choose \sigma = t_1 + t_2? why not some other function?
I will greatly appreciate if you can kindly refer me some reading on this topic.

The lines \tau= t_1+ t_2= constant is the same as t_2= -t_1+ constant have slope -1. The lines \sigma= t_1- t_2= constant or t_2= t_1- constant have slope 1. They are perpendicular so we still have an "orthogonal" coordinate system.

I have attached my solution as well. I have not introduced a new variable, however, I have used \tau and t_1.

Thank you for your time.