Two things define the Fourier transform — limits of −∞ to +∞, and a kernel of \(e^{-j\omega t}\) — and only option 4 has both.

\(F(\omega)=\int_{-\infty}^{+\infty}f(t)\,e^{-j\omega t}\,dt\)

Why the limits must be two-sided. The transform decomposes a signal into everlasting sinusoids, which have no beginning and no end, so the analysis must cover all time. Truncating the lower limit to zero, as options 1 and 2 do, silently assumes the signal is causal — and discards whatever happens for \(t\lt0\). That one-sided integral with the \(e^{-j\omega t}\) kernel is the unilateral transform, a different object.

Why the kernel carries a minus sign. The forward transform correlates \(f(t)\) against \(e^{+j\omega t}\), and correlation takes the complex conjugate — hence \(e^{-j\omega t}\). The positive exponent of option 3 is the kernel of the inverse transform:

\(f(t)=\dfrac{1}{2\pi}\int_{-\infty}^{+\infty}F(\omega)\,e^{+j\omega t}\,d\omega\)

Using the wrong sign yields \(F(-\omega)\) instead of \(F(\omega)\) — harmless for real even signals, wrong in general.

Option 1 is worth a second look because the two transforms are close relatives. Put \(s=\sigma+j\omega\) and set \(\sigma=0\): the Laplace kernel becomes the Fourier kernel. So the Fourier transform is the Laplace transform evaluated on the imaginary axis — valid whenever that axis lies inside the region of convergence. The extra \(e^{-\sigma t}\) is what lets Laplace handle growing signals that have no Fourier transform at all.

Convergence. The integral exists when \(f(t)\) is absolutely integrable; sinusoids and steps are not, and appear in the transform as impulses, admitted through the generalised-function extension of the theory.

Hence, the correct expression is the two-sided integral with kernel e^−jωt.