Why does Fourier transform gives the correct amplitudes?

Question

For instance we would like to get the Fourier transform of $A\cdot cos(2\pi fx)$. At some point, when we find the frequency $f$, we arrive to the following integral. $$\int_{-T}^{T}A\cdot cos^2(2\pi fx)dx=AT+\frac{sin(4\pi fT)}{4\pi f} \space ,\lim T=\infty$$ At this point I do not understand how does this give us the correct amplitude $A$ at the frequency $f$. The result seems to be infinity for me instead of $A$. I used the formula: $$\int_{-\infty}^{\infty}g(x)\cdot exp(-i2\pi fx)dx$$ Or did I misunderstand somehting and this formula gives infinity at the correct frequencies? I have noticed $AT+\frac{sin(4\pi fT)}{4\pi f}$ contains the amplitude $A$. If the frequency is high, the part $\frac{sin(4\pi fT)}{4\pi f}$ is small, so dividing by T which is known, we get close to $A$. But what happens when the frequency is low, or $\lim T=\infty$ ?