# The trick didn't like me (teaching Fourier transform)

I was teaching Fourier transform for engineering students. Since I didn't want to go into rigourous proofs during class, I often use intuition, just give students an idea to persuade them with the theory.

For example, if $$F(x):=\int_{-\infty}^{+\infty}\frac{\cos(wx)}{1+w^2}dw,$$ then we can differentiate $$F$$ by taking the derivative under the integral sign on the right hand side. For example, this is exactly what I did when I gave my students an explanation (idea) for the formula $$(\mathcal{F}(f))'(w)=-i\mathcal{F}(xf(x))$$.

The problem now is that, for the above integral (which we encounter a lot in the course), we can easily check that (using software) $$F(x)=\pi e^{-x},\forall x\in\mathbb{R},$$ on the other hand, if we differentiate under the integral sign, students then would expect that $$\pi e^{-x}=\int_{-\infty}^{\infty}\frac{\omega \sin(wx)}{1+w^{2}}dw,\forall x\in\mathbb{R}.$$ However, this is not true, as we can check that this is only true for $$x>0$$.

So the problem now is: well,... (I bit myself in front of engineering students, who were already terrified with Maths). I understand that in general differenting under the integral sign is not always legitimate.

How did you teach and go around this in your teaching? How to teach the formula: $$(\mathcal{F}(f))'(w)=-i\mathcal{F}(xf(x))$$

when students easily face and get confused with the above confusing matter?

Thanks.

• You should check your software. From the definition of $F$ you can see that $F(x)=F(-x)$,. – Dirk Nov 3 '19 at 13:44
• Sorry, I got a big mistake here. – Hana Puk Nov 3 '19 at 13:55
• If I look for errors in my argumentations, I usually start to look at the points where I wrote "easily"... – Dirk Nov 3 '19 at 14:47
• I believe F(x)=pi*e^-abs(x) – zokomoko Nov 10 '19 at 20:59

It's certainly beneficial for students (even if thy don't like maths) to be convinced that some manipulations are only valid under some conditions.

Maybe the following example is both significant and short enough for a rapid illustration:

Derivation under the integral sign is ("formally"): $$\frac{d}{dt} \int f(x,t) dx = \int \frac{\partial f(x,t)}{\partial t} dx$$ Derivation is essentially a limit: $$\frac{\partial f(x,t)}{\partial t} = \lim_{h\rightarrow 0}\frac{ f(x,t+h)-f(x,t)}{h}$$ Derivation under the integral sign consists of permuting the "limit" and "integration" operations: $$\lim \int = \int \lim$$ However this operation is not innocuous and is only valid under certain conditions, a counterexample being: $$\lim_{t\rightarrow 0}\int_0^\infty te^{-tx}dx=1\neq 0 = \int_0^\infty \lim_{t\rightarrow 0}te^{-tx}dx$$

• Nice example. In contrast, appreciating why cosine blows up is not too familiar unless the students either have taken or are taking a complex analysis class (apparently I need to take it again) – James S. Cook Nov 16 '19 at 4:12
• @JamesS.Cook I like this example too as it is very simple and easy to remember. I understand you comment. Maybe you can have a look at DIFFERENTIATING UNDER THE INTEGRAL SIGN an awesome document where your example is in section 13, but one need to go deeper into maths. – Picaud Vincent Nov 16 '19 at 6:45

Interesting integral. $$f(w)=\frac{\cos(wx)}{1+w^2}$$ has poles at $$w = \pm i$$. The residue at $$w=i$$ is given as follows: $$\text{Res}_{w=i}\left(\frac{\cos(wx)}{1+w^2}\right) = \text{Res}_{w=i}\left(\frac{\cos(wx)}{(w+i)(w-i)}\right) = \frac{\cos(ix)}{i+i} = \frac{\cosh(x)}{2i}$$ So, if I use the usual half-circle contour $$C = [-R,R] \cup C_R^+$$ where $$[-R,R]$$ is the line-segment on the real-axis connecting $$w=-R$$ and $$w=R$$ and $$C_R^+$$ is the CCW-oriented half-circle $$|w|=R$$ which goes from $$w=R$$ back to $$w=-R$$ then we note the Residue Theorem indicates: $$\frac{2\pi i\cosh(x)}{2i} = \int_{C} \frac{\cos(wx)dw}{1+w^2} = \int_{[-R,R]} \frac{\cos(wx)dw}{1+w^2}+\int_{C_R^+} \frac{\cos(wx)dw}{1+w^2}$$ as $$R \rightarrow \infty$$ you can not argue the half-circle integral vanishes as Paul Garret pointed out... hence I do not get to conclude $$\pi \cosh(x) = \int_{-\infty}^{\infty} \frac{\cos(wx)dw}{1+w^2}$$ because the half-circle integral does not nicely simplify in the limit.

• The integrals over the half-circles do not vanish, however. Also, qualitatively, since $1/(1+w^2)$ is of only modest decay, its Fourier transform will not be smooth, etc. – paul garrett Nov 15 '19 at 14:08
• @paulgarrett Yes, I should remember $\cos(wx)$ is unbounded, for example, if $x$ is real and $w=iR$ then $\cos(wx)=\cos(ixR)=cosh(xR)$ which blows up as $R \rightarrow \infty$. Curses. I will edit. – James S. Cook Nov 16 '19 at 4:05