Given the signal:
$ cos(4 \pi f)e^{-j 2 \pi 5 f} $
I'm trying to apply the inverse fourier transform like this:
$ \int^{+\infty}_{-\infty} cos(4 \pi f)e^{-j 2 \pi 5 f} \cdot e^{j 2 \pi f t} \, df $
$ \int^{+\infty}_{-\infty} \frac{e ^{-j 4 \pi f} + e ^ {j 4 \pi f}} {2} e^{-j 2 \pi 5 f} \cdot e^{j 2 \pi f t} \, df $
However, after applying the shift by splitting the cosine into two exponentials and multiplying those by $e^{j 2 \pi f t}$ I get stuck, I know that the result should be two separate impulses, but I don't know how to get there by integrating two exponentials.
So, first remark that since we are dealing here with generalized functions (these functions are not integrable, so the integral does not make sense with the usual definition) we should use tempered distributions to prove things rigorously. However, most of the properties are similar (and one should multiply by test functions to get them) so I will denote by $$ \mathcal{F}(u)(t) := \int_{-\infty}^∞ u(y)\,e^{-2iπ\,t\,y}\,\mathrm{d}y, $$ the Fourier transform and its extension to tempered distributions and $\mathcal{F}^{-1}$ its inverse. Then $$ \begin{align*} \mathcal{F}^{-1}(\cos(4\pi x)\,e^{-2iπ \,5x})(t) &= \frac{1}{2} \int_{-\infty}^∞ (e^{-2iπ \,3x} + e^{-2iπ \,7x})\, e^{-2iπ\,t\,y}\,\mathrm{d}y \\ &= \frac{1}{2} \int_{-\infty}^∞ e^{-2iπ \,(t-3)\,x} + e^{-2iπ \,(t-7)\,x}\,\mathrm{d}y \\ &= \frac{1}{2} \left(\mathcal{F}^{-1}(1)(t-3) + \mathcal{F}^{-1}(1)(t-7)\right) \\ &= \frac{1}{2} \left(\delta_0(t-3) + \delta_0(t-7)\right) \\ &= \frac{1}{2} \left(\delta_3(t) + \delta_7(t)\right). \end{align*} $$ I used the fact that the Fourier transform of the Dirac delta is the constant function $\mathcal{F}(\delta_0) = 1$, which follows by writting that for any $\varphi$ in the Schwartz class $$ \langle \mathcal{F}(\delta_0),\varphi\rangle = \langle \delta_0,\mathcal{F}(\varphi)\rangle = \mathcal{F}(\varphi)(0) = \int_{-\infty}^∞ \varphi = \langle 1,\varphi\rangle $$ and so $\mathcal{F}^{-1}(1) = \delta_0$.