Integral Euler's formula equals integral $\frac{sin(t)}{t}$ dt

Question

For real $c$ we should have that \begin{align} \int_{-T}^{T} \frac{e^{itc}}{2it} dt = \int_{0}^{T} \frac{\text{sin}(tc)} {t}dt. \end{align} However, for me this is not directly clear. I know that $e^{itc} = \text{cos}(tc) + i \text{sin}(tc)$ which result in \begin{align} \int_{-T}^{T} \frac{e^{itc}}{2it} dt &= \int_{-T}^{T}\frac{\text{cos}(tc) + i \text{sin}(tc)}{2it} dt\\ & = \frac{1}{2i} \int_{-T}^{T} \frac{\text{cos}(tc)}{t} dt + \int_{0}^{T} \frac{\text{sin}(tc)}{t} dt. \end{align} From here it has to follow that $\int_{-T}^{T} \frac{\text{cos}(tc)}{t} dt$ has to be equal to $0$. And this is not the case, is it?

Answer 1

First, you must interpret this as a principal value because

$$\int_{-T}^T\frac{e^{itc}}{2it}dt$$

passes through $t=0$, which is a singularity.

Secondly,

$$f(t)=\frac{\cos(tc)}{t}$$

I note that it is an odd function ($f(-t)=-f(t)$), so if we look at this integral, we see that

$$\int_{-T}^Tf(t)dt=\int_{-T}^0f(t)dt+\int_0^Tf(t)dt$$

$$=\int_0^Tf(-t)dt+\int_0^Tf(t)dt$$

$$=-\int_0^Tf(t)dt+\int_0^Tf(t)dt$$

If we have $\int_0^Tf(t)dt=F(T)$, then

$$=-F(T)+F(T)=0$$

Of course, you must also note that $F(T)\to\infty$, but ignoring that, it equals $0$.