# Double integral computation

1. Feb 5, 2014

### inviziblesoul

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

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2. Feb 5, 2014

### HallsofIvy

Staff Emeritus
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
$$\int_{-2T}^0\int_{2T-\tau}^{\tau+ 2T} c(\tau)d\sigma d\tau+ \int_0^{2T}\int_{-\tau- 2T}^{\tau- 2T} c(\tau) d\sigma d\tau$$

3. Feb 5, 2014

### 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 $t_1$.

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• ###### DOC050214-05022014164355.pdf
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4. Feb 6, 2014

### HallsofIvy

Staff Emeritus
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.

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.