double integral to simple integral this question is probably simple but i just can't find out how to change this double integral to simple integral. Can you please help me? :)
Thanks very much for your help!
f is a continuous/C1 function (verifying all the hypothesis to be integrable)

 A: 
The relationship in the highlighted box of the OP is not generally true.  For example, if $f$ is an odd function then $\int_0^T\int_0^T f(t-u)\,dt\,du=0$ while $2\int_0^T(T-x)f(x)\,dx$ need not be $0$ when $f$ is odd (Take $f(x)=x$, for example).

However, we can assert that the relationship holds for all even functions.  To show this we will write $f(x)=f^e(x)+f^o(x)$ where $f^e$ and $f^o$ are the even and odd components of $f$, respectively and given by 
$$f^e(x)=\frac12 (f(x)+f(-x))\\\\f^o(x)=\frac12 (f(x)-f(-x))$$
Next, enforcing the substitution $x=t-u$ in the inner integral yields
$$\begin{align}
\int_0^T\int_0^T f(t-u)\,dt\,du&\overbrace{=}^{x=t-u}\int_0^T\int_{-u}^{T-u}f(x)\,dx\,du\\\\
&=\int_0^T \int_0^{T-x}f(x)\,du\,dx+\int_{-T}^0\int_{-x}^T f(x)\,du\,dx\\\\
&=\int_0^T f(x) \int_0^{T-x}(1)\,du\,dx+\int_{-T}^0 f(x)\int_{-x}^T(1)\,du\,dx\\\\
&=\int_0^T (T-x)f(x)\,dx+\int_{-T}^0(T+x)f(x)\,dx\\\\
&=\int_0^T (T-x)(f(x)+f(-x))\,dx\\\\
&=2\int_0^T(T-x)f^e(x)\,dx
\end{align}$$
Hence, we find that 

$$\int_0^T\int_0^T f(t-u)\,dt\,du=2\int_0^T(T-x)f^e(x)\,dx\tag 1$$

Note that we could have written 
$$\begin{align}
\int_0^T\int_0^T f(t-u)\,dt\,du&=\int_0^T\int_0^T f^e(t-u)\,dt\,du+\overbrace{\int_0^T\int_0^T f^o(t-u)\,dt\,du}^{=0}\\\\
&=\int_0^T\int_0^T f^e(t-u)\,dt\,du\tag 2
\end{align}$$
Combining $(1)$ and $(2)$ yields

$$\int_0^T\int_0^T f^e(t-u)\,dt\,du=2\int_0^T(T-x)f^e(x)\,dx$$

