Why is it wrong to substitute $u=1/t$ in $\int_{-\infty}^{\infty}f\left(t\right)dt$ and get an integral from $0$ to $0$?

Question

As the title says, suppose we have an integral: $$\int_{-\infty}^{\infty}f\left(t\right)dt,$$ and we set $t=\frac{1}{u}$. Then, we get $$-\int_{0}^{0}f\left(\frac{1}{u}\right)\frac{1}{u^{2}}du=0.$$

Where did this go wrong?

Answer 1

It does not work because $\frac{1}{t}$ is not defined at $0$. However, you can split the integral in two parts $$ \int_{-\infty}^{+\infty}f(t)dt=\int_{-\infty}^0f(t)dt+\int_0^{+\infty}f(t)dt $$ and set $u=\frac{1}{t}$ in each integral ($\frac{1}{t}$ is $\mathscr{C}^1$ on $(-\infty,0)$ and $(0,+\infty)$ !). You get $$ \int_{-\infty}^{+\infty}f(t)dt=\int_{-\infty}^0\frac{f(1/u)}{u^2}du+\int_0^{+\infty}\frac{f(1/u)}{u^2}du=\int_{-\infty}^{+\infty}\frac{f(1/u)}{u^2}du $$