Prove that $f\notin C^1(\Bbb R)$ knowing that $f(t)=\int_{-\infty}^{\infty} \frac{e^{2i\pi tx}}{1+x^2}\ dx$ The goal here is to prove that $f\notin C^1(\Bbb R)$ knowing that $f(t)=\int_{-\infty}^{\infty} \frac{e^{2i\pi tx}}{1+x^2}\ dx$. I am really confused about this one as later on I show using the residues theorem (that I'm told to assume I can use) that, if $g(x) = \frac{e^{2i\pi tx}}{1+x^2}$ :
$$f(t) = \sum_{k=1}^{n}2i\pi \ res(h,z_k)$$
 for residues whose associated poles have an imaginary part $>0$. As $i$ is the only such pole we get :
$$f(t) = 2i\pi \ res(h,z_k)$$ with $$res(h,z_k)=\frac{1}{2i}e^{-2\pi t}$$ and so
$$f(t) = \pi e^{-2\pi t}$$ which is actually $C^1(\Bbb R)$.
Is my use of the residues theorem wrong, or is there a mistake in the question (i.e. $\notin$ instead of $\in$)? Any advice would be appreciated!
 A: It should be $\pi\exp(-2\pi|t|)$, since the sign of $t$ affects which semicircular contour should enclose a pole.
A: We've seen that one can calculate $f(t)$ explicitly, and it turns out that $f$ is not differentiable at the origin. One can also show this without actually calculating $f(t)$. Why might we want to do this? (i) the argument below is more fun (ii) I think if you include all the details it's simpler (iii) most important, what's below is the sort of thing one can imagine applying in some other context, where one cannot evaluate the integral explicitly.
It's clear that $f$ is even, so if $f$ is differentiable at the origin we must have $f'(0)=0$. So it's enough to show that $$\lim_{t\to0^+}\frac{f(t)-f(0)}t\ne0.$$
If $t>0$, a change of variables $x'=tx$ (plus the fact that the integral of a function over the line is the integral of its even part) shows that $$\frac{f(t)-f(0)}t=\int\frac{\cos(2\pi x)-1}{x^2+t^2}\,dx.$$
The integrand is bounded by $c$ for $|x|\le1$ and by $2/x^2$ for $|x|>1$, so DCT shows that $$\lim_{t\to0^+}\frac{f(t)-f(0)}t=\int\frac{\cos(2\pi x)-1}{x^2}\,dx<0$$(because $\cos(2\pi x)-1<0$ almost everywhere).
(???: We said we were going to show that $f$ is not differentiable at the origin, but above we appear to have shown that $f'(0)$ equals that (finite) integral. Huh? No, actually all we've shown is that the right hand derivative of $f$ at the origin is given by that integral. Exercise: Show that the left-hand derivative  at the origin is a slightly different integral.)
