In the proofs of the convolution theorem I have seen, say here, there is a change of variable used in a nested integral. Specifically, within an outer integral with respect to x, there is an inner integral with respect to u, independent of x; and for which the substitution $w=u-x$ is made. Then $dw$ is supposedly equal to $du$. I don't quite understand why we don't have $dw=du-dx$, considering that u and x are independent. I imagine that someone may motivate this by saying something like 'within the second integral, x is a constant', and I can kind of get this by thinking about integrals as sums, and indeed in the inner summation we would perform it for each value of x (in the outer integral). However I am not convinced by this, and was wondering if anyone might have a better explanation, or some more rigorous motivation (although I'm not sure i'll understand something very rigorous!!!)
[Math] Change of variable in proof of convolution theorem
change-of-variableconvolutionfourier transformintegration
Related Solutions
Yes, you can find it directly as:
$$\displaystyle \int_{-\infty}^{\infty} e^{- \tau}~u(\tau)~e^{-2(t-\tau)}~u(t-\tau)~d\tau = e^{-2t}~\int_0^t e^{\tau}~d\tau = e^{-2t}(e^t-1)$$
A plot shows:
When we have functions $f(t)u(t)$ and $g(t)u(t)$ with the Heaviside Unit Step Function, we can just write:
$$\displaystyle (f*g)(t) = \int_0^t f(\tau)~g(t-\tau) ~ d\tau$$
Having said all of that, I think it is very important to understand what is going on graphically. I recommend spending time with the examples, particulary $3.4.1$, Example $1$ as they solve the general example to yours and do it both ways. It is critical to understand the graphical method as it can keep you away from unrecognizable integrals.
This is also a useful Convolution Table. Especially review "Convolution using graphical method (1)".
Found the solution on: Fourier Transform Proof $ \mathcal F(f(x)g(x))=(\frac{1}{2\pi})F(s)*G(s)$
Translated to our notation, this would give:
Starting with the claim $\widehat{f(x)g(x)}=(\hat{f}*\hat{g})(\xi)$ and applying the FT inversion formula on both sides we obtain: $$f(x)g(x)=\int_{\mathbb{R}^d}e^{2\pi i x\cdot\xi}(\hat{f}*\hat{g})(\xi)d\xi$$ Now with the definition of the convolution we get: $$f(x)g(x)=\int_{\mathbb{R}^d}e^{2\pi i x\cdot\xi}\bigg(\int_{\mathbb{R}^d}\hat{f}(\zeta)\hat{g}(\xi-\zeta)d\zeta\bigg)d\xi$$ multiplying the RHS with $1=e^{2\pi i x\cdot \zeta}e^{-2\pi i x\cdot \zeta}$ we have: $$f(x)g(x)=\int_{\mathbb{R}^d}e^{2\pi i x\cdot\xi}\int_{\mathbb{R}^d}\hat{f}(\zeta)\hat{g}(\xi-\zeta)e^{2\pi i x\cdot\zeta}e^{-2\pi i x\cdot \zeta}d\zeta d\xi$$ Because $|\phi(z)e^{az}|=|\phi(z)|$ we can use Fubini on the integrals above, which gives us: $$f(x)g(x)=\int_{\mathbb{R}^d}e^{2\pi i x\cdot\zeta}\hat{f}(\zeta)d\zeta\int_{\mathbb{R}^d}e^{2\pi i x\cdot(\xi-\zeta)}\hat{g}(\xi-\zeta)d\xi$$ from which we finally deduce the true statement: $$f(x)g(x)=f(x)g(x)$$ hence our starting equation $\widehat{fg}=\hat{f}*\hat{g}$ is true and we are done.
Best Answer
$$ h(u) = \int_{-\infty}^\infty f(x) g(u-x)\,dx. $$
In the expression $\displaystyle \int_{-\infty}^\infty \cdots\cdots\,dx,$ anything that remains fixed as $x$ goes from $-\infty$ to $+\infty$ is a “constant.” Thus $\dfrac d{dx} (u-x) = 0-1,\vphantom{\dfrac{\displaystyle\sum}{}}$ so if we set $y=u-x,$ then $dy = -dx.$