Making double integral into single one

Question

How can we show that

$\int^{\frac{T}{2}}_{-\frac{T}{2}}ds\int^{\frac{T}{2}}_{-\frac{T}{2}}dt\ g(s-t) = \int^{T}_{-T}d\tau\ g(\tau)(T-|\tau|)$ ?

Seems we need to set $\tau = s-t$, but I don't know how to set the other variable.

Answer 1

The answer is: set the other variable as $s+t$.

See below for details.

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Consider $s$ as the horizontal coordinate and $t$ as the vertical coordinate.

As you have figured out, set $\tau = s - t$, which contour lines (constant level lines where $\tau = s - t =k$ has a fixed value) are going diagonally (northeast-southwest). Namely, the "axis" for $\tau$ points in the northwest-southeast direction (just like how $x$-axis goes horizontally while the lines of constant $x$ is vertical).

Set the other variable $\sigma = s+t$ which contour is perpendicular to $\tau$. The transformation from $\{s,t\}$ to $\{\sigma,\, \tau\}$ will give you a scaling factor (Jacobian) of $1/\sqrt{2}$ that needs be multiplied.

The integrand $g(s,t) = g(s-t) = g(\tau)$ is a function of $\tau$ only. That is, it is constant along the northeast-southwest direction, for any given $\sigma$.

The given integration limit of $\{s,t\} \in [\frac{T}2, \frac{T}2]^2$ is a square of side length $T$ where the diagonal is of length $\sqrt{2}T$. In the new frame (coordinates with $\sigma, \tau$) it is diamond shaped.

The infinitesimally thin strip of a given $\sigma$ where $g(\tau)$ is constant is of length $\sqrt{2}(T - |\tau|)$. This is where that factor $(T - |\tau|)$ comes from, after multiplying with the aforementioned Jacobian. The absolute value on $|\tau|$ is due to the fact that we are talking about length (magnitude) and that the diamond is symmetric.