Change of variables in proof of Bochner's theorem I have $\phi: \mathbb{R} \to \mathbb{R}$ continuous, bounded, and integrable. I'm going through a book which makes the following calculation:
$$\frac{1}{2\pi}\int\limits_{-\infty}^{\infty}e^{-itx}\phi(t)dt = \lim\limits_{T\to \infty}
\frac{1}{2\pi}\int\limits_{-T}^{T}\left(1 - \frac{|t|}T \right)e^{-itx}\phi(t)dt = 
\lim\limits_{T\to \infty}
\frac{1}{2\pi T}\int\limits_{0}^{T} \int\limits_{0}^{T} e^{-i(t-s)x}\phi(t-s)dtds.
$$ 
The first equality holds by the Dominated Convergence Theorem and the second is supposed to hold by a change of variables. 
Can someone please show me how this change of variables goes?
Edit: If it helps, this is in the book of Varadhan in the proof of Bochner's theorem. This computation is being made to use the positive-definiteness of $\phi$.
 A: Let's prove it backwards. We have a function $A(y)=e^{-iyx}\phi(y)$ which we want to integrate over a region.
$$
\int_0^T \int_0^T A(t-s)dt ds
$$
The integrating region is a square $[0,T]\times [0,T]$, however our function has certain symmetry, it is not any function in two variables $t,s$ but a function in one variable $t-s$. This means that it is constant on level curves of $C(t,s)=t-s$, i.e., on lines of slope ${45}^\circ$.
Let's introduce a change of variables
$$
u = t - s\\
v = t + s
$$
We have $dtds =\tfrac{1}{2} dudv$.
The variables are constrained by $u\in[-T,T]$ and $v\in [|u|,2T-|u|]$, you can see this by drawing the region over which you integrate. Therefore after change of variables we obtain
$$
\int_0^T \int_0^T A(t-s)dt ds = \int_{-T}^T \int_{|u|}^{2T-|u|}A(u) \tfrac{1}{2}dvdu =\\
\int_{-T}^T \left(\int_{|u|}^{2T-|u|} \tfrac{dv}{2}\right )A(u)du = \int_{-T}^T (\tfrac{2T - 2|u|}{2})A(u)du = T\int_{-T}^T\left(1-\frac{|u|}{T}\right) A(u)du.
$$
Plugging in the correspoiding limits and constant, dividing everything by $T$ and renaming $u$ to $t$ gives the full solution.
A: Let's see this integration as a set of convolutions. In fact, if we set
\begin{align}
q(t) &= u(t)-u(t-T),\\
q'(t)&=q(-t)
\end{align}
where $u(\cdot)$ is the heaviside step function, and
\begin{equation}
p(t)= {\rm e}^{-itx} \phi(t).
\end{equation}
Hence, one can easily check that
\begin{equation}
\Big(1-\frac{|t|}{T}\Big) \big(u(t+T)-u(t-T)\big) = \frac{1}{T}(q'*q)(t),
\end{equation}
in which $*$ is the convolution sign. As a result, one can see that
\begin{equation}
\int_{-T}^T \Big(1-\frac{|t|}{T}\Big) {\rm e}^{-itx} \phi(t){\rm d} t = \frac{1}{T}(q'*q*p)(0).
\end{equation}
Next, using the definiton of $q(t)$, you can rewrite the above formula to see that
\begin{equation}
(q*p)(t) = \int_{s=-\infty}^{\infty} \big(u(s)-u(s-T)\big) p(t-s){\rm d} s = \int_{s=0}^{T} p(t-s){\rm d} s.
\end{equation} 
Finally, by associativity of convolution, we have that
\begin{equation}
\frac{1}{T}(q'*q*p)(0) = \int_{t=0}^T \int_{s=0}^{T} p(t-s){\rm d} s{\rm d} t,
\end{equation}
which completes the proof.
A: Let $s\in\mathbb{R}$ (fixed). Start by $\displaystyle{\frac{1}{2\pi}\int_{-T}^T\bigg{(}1-\frac{|t|}{T}\bigg{)}e^{-itx}\phi(t)dt}$ and change the variable to $t=t'-s$. Then you get the expression $\displaystyle{\frac{1}{2\pi}\int_{-T-s}^{T-s}\bigg{(}1-\frac{|t-s|}{T}\bigg{)}e^{-i(t-s)x}\phi(t-s)dt}$ (note that since $t'$ is a dummy variable i simply wrote $t$ instead, its the same thing). This is true for any $s\in\mathbb{R}$, so as a function of $s$, this is constant. Integrate with respect to $s$ over $[0,T]$; thus this integral is equal to $T\cdot\frac{1}{2\pi}\int_{-T}^T(1-\frac{|t|}{T})e^{-itx}\phi(t)dt$ (we simply integrated a constant function over an interval!), so we have
$$ \frac{1}{2\pi}\int_{-T}^T\bigg{(}1-\frac{|t|}{T}\bigg{)}e^{-itx}\phi(t)dt=\frac{1}{2\pi T}\int_0^T\int_{-T-s}^{T-s}\bigg{(}1-\frac{|t-s|}{T}\bigg{)}e^{-i(t-s)x}\phi(t-s)dtds$$
Now I am almost sure that breaking the inside integral at $s$ to lose the absolute value will yield the desired expression. EDIT: apparently this breaking the integral won't work. But draw the image of the region of the plane over which you are integrating. it is a parallelogram with edges $(T,0), (-T,0), (-2T,T), (0,T)$. Look for a linear transformation that will transform this to the square $[0,T]\times [0,T]$ and apply this as a change for variables. This should work.
