f^(ω)=F{f(t)}=∫−∞∞f(t)e−iωtdt{\displaystyle {\hat {f}}(\omega )={\mathcal {F}}{f(t)}=\int _{-\infty }^{\infty }f(t)e^{-i\omega t}\mathrm {d} t}

f(t)=F−1{f^(ω)}=12π∫−∞∞f^(ω)eiωtdω{\displaystyle f(t)={\mathcal {F}}^{-1}{{\hat {f}}(\omega )}={\frac {1}{2\pi }}\int _{-\infty }^{\infty }{\hat {f}}(\omega )e^{i\omega t}\mathrm {d} \omega }

F{f′(t)}=∫−∞∞f′(t)e−iωtdt,  u=e−iωt, v=f′(t)dt=iωf^(ω){\displaystyle {\begin{aligned}{\mathcal {F}}{f^{\prime }(t)}&=\int _{-\infty }^{\infty }f^{\prime }(t)e^{-i\omega t}\mathrm {d} t,\ \ u=e^{-i\omega t},\ v=f^{\prime }(t)\mathrm {d} t\&=i\omega {\hat {f}}(\omega )\end{aligned}}} In general, we can take n{\displaystyle n} derivatives. F{f(n)(t)}=(iω)nf^(ω){\displaystyle {\mathcal {F}}{f^{(n)}(t)}=(i\omega )^{n}{\hat {f}}(\omega )} This yields the interesting property, stated below, which may be familiar in quantum mechanics as the form that the momentum operator takes in position space (on the left) and momentum space (on the right). [5] X Research source −iddt→ω{\displaystyle -i{\frac {\mathrm {d} }{\mathrm {d} t}}\to \omega }

F{tf(t)}=∫−∞∞tf(t)e−iωtdt=∫−∞∞i∂∂ω(e−iωt)f(t)dt=iddωf^(ω){\displaystyle {\begin{aligned}{\mathcal {F}}{tf(t)}&=\int _{-\infty }^{\infty }tf(t)e^{-i\omega t}\mathrm {d} t\&=\int _{-\infty }^{\infty }i{\frac {\partial }{\partial \omega }}(e^{-i\omega t})f(t)\mathrm {d} t\&=i{\frac {\mathrm {d} }{\mathrm {d} \omega }}{\hat {f}}(\omega )\end{aligned}}} In general, we can multiply by tn. {\displaystyle t^{n}. } F{tnf(t)}=indndωnf^(ω){\displaystyle {\mathcal {F}}{t^{n}f(t)}=i^{n}{\frac {\mathrm {d} ^{n}}{\mathrm {d} \omega ^{n}}}{\hat {f}}(\omega )} We immediately obtain the below result. This is a symmetry that is not fully realized with the Laplace transforms between the variables t{\displaystyle t} and s. {\displaystyle s. } iddω→t{\displaystyle i{\frac {\mathrm {d} }{\mathrm {d} \omega }}\to t}

F{eiatf(t)}=∫−∞∞f(t)e−i(ω−a)tdt=f^(ω−a){\displaystyle {\mathcal {F}}{e^{iat}f(t)}=\int _{-\infty }^{\infty }f(t)e^{-i(\omega -a)t}\mathrm {d} t={\hat {f}}(\omega -a)}

F{f(t−c)}=∫−∞∞f(t−c)e−iωtdt=∫−∞∞f(t)e−iω(t+c)dt=e−iωcf^(ω){\displaystyle {\begin{aligned}{\mathcal {F}}{f(t-c)}&=\int _{-\infty }^{\infty }f(t-c)e^{-i\omega t}\mathrm {d} t\&=\int _{-\infty }^{\infty }f(t)e^{-i\omega (t+c)}\mathrm {d} t\&=e^{-i\omega c}{\hat {f}}(\omega )\end{aligned}}}

F{f(ct)}=∫−∞∞f(ct)e−iωtdt,u=ct=1|c|∫−∞∞f(u)e−iωu/cdu=1|c|f^(ωc){\displaystyle {\begin{aligned}{\mathcal {F}}{f(ct)}&=\int _{-\infty }^{\infty }f(ct)e^{-i\omega t}\mathrm {d} t,\quad u=ct\&={\frac {1}{|c|}}\int _{-\infty }^{\infty }f(u)e^{-i\omega u/c}\mathrm {d} u\&={\frac {1}{|c|}}{\hat {f}}\left({\frac {\omega }{c}}\right)\end{aligned}}}

F{f(t)∗g(t)}=∫−∞∞e−iωtdt∫−∞∞f(t−y)g(y)dy,u=t−y=∫−∞∞e−iω(u+y)du∫−∞∞f(u)g(y)dy=∫−∞∞f(u)e−iωudu∫−∞∞g(y)e−iωydy=f^(ω)g^(ω){\displaystyle {\begin{aligned}{\mathcal {F}}{f(t)*g(t)}&=\int _{-\infty }^{\infty }e^{-i\omega t}\mathrm {d} t\int _{-\infty }^{\infty }f(t-y)g(y)\mathrm {d} y,\quad u=t-y\&=\int _{-\infty }^{\infty }e^{-i\omega (u+y)}\mathrm {d} u\int _{-\infty }^{\infty }f(u)g(y)\mathrm {d} y\&=\int _{-\infty }^{\infty }f(u)e^{-i\omega u}\mathrm {d} u\int _{-\infty }^{\infty }g(y)e^{-i\omega y}\mathrm {d} y\&={\hat {f}}(\omega ){\hat {g}}(\omega )\end{aligned}}}

The Fourier transform of an even function fe(t){\displaystyle f_{e}(t)} is also even, because the integral is even in ω{\displaystyle \omega } due to the cos⁡ωt. {\displaystyle \cos \omega t. } Furthermore, if fe(t){\displaystyle f_{e}(t)} is real, then its Fourier transform is also real. F{fe(t)}=∫−∞∞fe(t)(cos⁡ωt−isin⁡ωt)dt=2∫0∞fe(t)cos⁡ωtdt{\displaystyle {\begin{aligned}{\mathcal {F}}{f_{e}(t)}&=\int {-\infty }^{\infty }f{e}(t)\left(\cos \omega t-i\sin \omega t\right)\mathrm {d} t\&=2\int {0}^{\infty }f{e}(t)\cos \omega t\mathrm {d} t\end{aligned}}} The Fourier transform of an odd function fo(t){\displaystyle f_{o}(t)} is also odd, because the integral is odd in ω{\displaystyle \omega } due to the sin⁡ωt. {\displaystyle \sin \omega t. } Furthermore, if fo(t){\displaystyle f_{o}(t)} is real, then its Fourier transform is purely imaginary. F{fo(t)}=∫−∞∞fo(t)(cos⁡ωt−isin⁡ωt)dt=−2i∫0∞fo(t)sin⁡ωtdt{\displaystyle {\begin{aligned}{\mathcal {F}}{f_{o}(t)}&=\int {-\infty }^{\infty }f{o}(t)\left(\cos \omega t-i\sin \omega t\right)\mathrm {d} t\&=-2i\int {0}^{\infty }f{o}(t)\sin \omega t\mathrm {d} t\end{aligned}}}

F{1t2+1}=∫−∞∞e−iωtt2+1dt{\displaystyle {\mathcal {F}}\left{{\frac {1}{t^{2}+1}}\right}=\int _{-\infty }^{\infty }{\frac {e^{-i\omega t}}{t^{2}+1}}\mathrm {d} t}

To use residues, we create a contour γ{\displaystyle \gamma } consisting of a concatenation of the real line and a semicircular arc in the lower half plane that circles clockwise. The goal is to show that the real integral equals the contour integral by showing that the arc integral vanishes. ∮γe−iωtt2+1dt=∫−∞∞e−iωtt2+1dt+∫arce−iωtt2+1dt{\displaystyle \oint _{\gamma }{\frac {e^{-i\omega t}}{t^{2}+1}}\mathrm {d} t=\int {-\infty }^{\infty }{\frac {e^{-i\omega t}}{t^{2}+1}}\mathrm {d} t+\int {\text{arc}}{\frac {e^{-i\omega t}}{t^{2}+1}}\mathrm {d} t} We may factor the denominator to show that the function has simple poles at t±=±i. {\displaystyle t{\pm }=\pm i. } Since only t−{\displaystyle t{-}} is being enclosed, we can use the residue theorem to calculate the value of the contour integral. Res⁡(f(t);−i)=e−ω−2i{\displaystyle \operatorname {Res} (f(t);-i)={\frac {e^{-\omega }}{-2i}}} Note that since our contour is in the clockwise direction, there is an additional negative sign. ∮γe−iωtt2+1dt=−2πi⋅e−ω−2i=πe−ω{\displaystyle \oint _{\gamma }{\frac {e^{-i\omega t}}{t^{2}+1}}\mathrm {d} t=-2\pi i\cdot {\frac {e^{-\omega }}{-2i}}=\pi e^{-\omega }} Equally important is the process in showing that the arc integral vanishes. Jordan’s lemma aids in this evaluation. While the lemma does not say that the integral vanishes, it does bound the difference between the contour integral and the real integral. [11] X Research source www. damtp. cam. ac. uk/user/reh10/lectures/nst-mmii-chapter5. pdf We apply the lemma to the lower half plane below for a function f(t)=e−iωtg(t),{\displaystyle f(t)=e^{-i\omega t}g(t),} where ω>0. {\displaystyle \omega >0. } Given a parameterization C=Re−iϕ{\displaystyle C=Re^{-i\phi }} where ϕ∈[0,π],{\displaystyle \phi \in [0,\pi ],} then Jordan’s lemma prescribes the following bound of the integral: |∫Cf(t)dt|≤πωmaxϕ∈[0,π]g(Re−iϕ){\displaystyle {\Bigg |}\int _{C}f(t)\mathrm {d} t{\Bigg |}\leq {\frac {\pi }{\omega }}\max _{\phi \in [0,\pi ]}g(Re^{-i\phi })} Now, all we need to do is show that g(t){\displaystyle g(t)} vanishes in the large R{\displaystyle R} limit, which is trivial here because the function falls off as 1/R2. {\displaystyle 1/R^{2}. } limR→∞1(Re−iϕ)2+1=0{\displaystyle \lim _{R\to \infty }{\frac {1}{(Re^{-i\phi })^{2}+1}}=0} What is the domain of ω{\displaystyle \omega } in this result? As stated previously, Jordan’s lemma only applies for ω>0. {\displaystyle \omega >0. } However, when one repeats this calculation by enclosing the upper half plane, finding the residue at the other pole, and applying Jordan’s lemma again to ensure the arc integral vanishes, the result will be πeω{\displaystyle \pi e^{\omega }} while the domain of ω{\displaystyle \omega } will be the negative reals. So the final answer is written below. F{1t2+1}=πe−|ω|{\displaystyle {\mathcal {F}}\left{{\frac {1}{t^{2}+1}}\right}=\pi e^{-|\omega |}}

F{rect⁡(t)}=∫−1/21/2e−iωtdt=1iω(eiω/2−e−iω/2)=2ωsin⁡ω2{\displaystyle {\begin{aligned}{\mathcal {F}}{\operatorname {rect} (t)}&=\int _{-1/2}^{1/2}e^{-i\omega t}\mathrm {d} t\&={\frac {1}{i\omega }}\left(e^{i\omega /2}-e^{-i\omega /2}\right)\&={\frac {2}{\omega }}\sin {\frac {\omega }{2}}\end{aligned}}} If the unit pulse is shifted such that the bounds are 0 and 1, then there exists an imaginary component as well, as seen by the graph above. This is due to the fact that the function is no longer even. ∫01e−iωtdt=sin⁡ωω+i(cos⁡ω−1ω){\displaystyle \int _{0}^{1}e^{-i\omega t}\mathrm {d} t={\frac {\sin \omega }{\omega }}+i\left({\frac {\cos \omega -1}{\omega }}\right)}

F{e−t2}=∫−∞∞e−t2e−iωtdt=∫−∞∞e−(t2+iωt−ω2/4+ω2/4)dt=e−ω2/4∫−∞∞e−(t+iω/2)2dt=πe−ω2/4{\displaystyle {\begin{aligned}{\mathcal {F}}{e^{-t^{2}}}&=\int _{-\infty }^{\infty }e^{-t^{2}}e^{-i\omega t}\mathrm {d} t\&=\int _{-\infty }^{\infty }e^{-(t^{2}+i\omega t-\omega ^{2}/4+\omega ^{2}/4)}\mathrm {d} t\&=e^{-\omega ^{2}/4}\int _{-\infty }^{\infty }e^{-(t+i\omega /2)^{2}}\mathrm {d} t\&={\sqrt {\pi }}e^{-\omega ^{2}/4}\end{aligned}}}

F{eiat}=2πδ(ω−a){\displaystyle {\mathcal {F}}{e^{iat}}=2\pi \delta (\omega -a)} The imaginary exponential oscillates around the unit circle, except when t=0,{\displaystyle t=0,} where the exponential equals 1. You can think of the contributions by the oscillations as canceling themselves out for all t≠0. {\displaystyle t\neq 0. } At t=0,{\displaystyle t=0,} the integral of the function then diverges. The delta function is then used to model this behavior. This result gives us the Fourier transform of three other functions for “free. " The Fourier transform of the constant function is obtained when we set a=0. {\displaystyle a=0. } F{1}=2πδ(ω){\displaystyle {\mathcal {F}}{1}=2\pi \delta (\omega )} The Fourier transform of the delta function is simply 1. F{δ(t)}=1{\displaystyle {\mathcal {F}}{\delta (t)}=1} Using Euler’s formula, we get the Fourier transforms of the cosine and sine functions. [13] X Research source F{cos⁡at}=π(δ(ω−a)+δ(ω+a)){\displaystyle {\mathcal {F}}{\cos at}=\pi (\delta (\omega -a)+\delta (\omega +a))} F{sin⁡at}=−iπ(δ(ω−a)−δ(ω+a)){\displaystyle {\mathcal {F}}{\sin at}=-i\pi (\delta (\omega -a)-\delta (\omega +a))}

F{tneiat}=2πindndωnδ(ω−a){\displaystyle {\mathcal {F}}{t^{n}e^{iat}}=2\pi i^{n}{\frac {\mathrm {d} ^{n}}{\mathrm {d} \omega ^{n}}}\delta (\omega -a)}

F{Θ(t)}=∫0∞e−iωtdt=e−iωt−iω|0∞=1iω{\displaystyle {\mathcal {F}}{\Theta (t)}=\int {0}^{\infty }e^{-i\omega t}\mathrm {d} t={\frac {e^{-i\omega t}}{-i\omega }}{\Bigg |}{0}^{\infty }={\frac {1}{i\omega }}} In order to make sense of this answer, we appeal to convolutions. The derivative of a convolution of two functions is given below. Note that this is not the product rule of ordinary derivatives. ddt(f(t)∗g(t))=f′∗g=f∗g′{\displaystyle {\frac {\mathrm {d} }{\mathrm {d} t}}(f(t)g(t))=f’g=fg’} Then, we see that the convolution of the derivative of an absolutely integrable function f(t){\displaystyle f(t)} with Θ(t){\displaystyle \Theta (t)} can be written in the following manner. This also implies the important relation Θ′(t)=δ(t). {\displaystyle \Theta ‘(t)=\delta (t). } f′(t)∗Θ(t)=∫−∞∞f′(y)Θ(t−y)dy=∫−∞tf′(y)dy=f(t){\displaystyle {\begin{aligned}f’(t)\Theta (t)&=\int _{-\infty }^{\infty }f’(y)\Theta (t-y)\mathrm {d} y\&=\int _{-\infty }^{t}f’(y)\mathrm {d} y=f(t)\end{aligned}}} F{f′(t)∗Θ(t)}=f^(ω)=f^′(ω)Θ^(ω)=iωf^(ω)Θ^(ω){\displaystyle {\mathcal {F}}{f’(t)*\Theta (t)}={\hat {f}}(\omega )={\hat {f}}’(\omega ){\hat {\Theta }}(\omega )=i\omega {\hat {f}}(\omega ){\hat {\Theta }}(\omega )} In this sense, we may then conclude that F{Θ(t)}=1iω. {\displaystyle {\mathcal {F}}{\Theta (t)}={\frac {1}{i\omega }}. }