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Calculus Cheatsheet

Common Integrals

Note: Constant of integration +C+C and trivial domain restrictions (usually a0a \ne 0) omitted for brevity except in places of particular significance.

(ax+b)ndx=(ax+b)n+1a(n+1),n1; x0, if n<01ax+bdx=1alnax+b+C1ax+bdx=1alnK(ax+b),C=1alnK; ax+b0\begin{alignat*}{2} &\int{\parens{ax + b}^n \,\diff{x}} = \frac{\parens{ax + b}^{n+1}}{a \parens{n+1}} ,\qquad &&\Exn{ %a \ne 0 n \neq -1 ;\ x \neq 0 ,\ \text{if } n < 0 } \\[1.4em] &\int{\frac{1}{ax + b} \,\diff{x}} = \frac{1}{a} \ln{\abs{ax + b}} + C \qquad && \\ &\phantom{\int{\frac{1}{ax + b} \,\diff{x}}} = \frac{1}{a} \ln{\abs{K(ax + b)}} ,\qquad &&\Exn{ C = \frac{1}{a} \ln{K} %;\ a \ne 0 ;\ ax + b \ne 0 } \end{alignat*}

eaxdx=1aeax \int{e^{ax} \,\diff{x}} = \frac{1}{a} e^{ax}

axdx=axlna,a1 \int{a^x \,\diff{x}} %= \int{\parens{e^{\ln{a}}}^x \,\diff{x}} % Derived from change of base law = \frac{a^x}{\ln{a}} ,\quad \Exn{a \neq 1}

cosaxdx=1asinaxsinaxdx=1acosaxsec2axdx=1atanaxtanaxdx=1alnsecax\begin{alignat*}{3} &\int{\cos{ax} \,\diff{x}} &&= && \frac{1}{a} \sin{ax} \\ &\int{\sin{ax} \,\diff{x}} &&= -&& \frac{1}{a} \cos{ax} \\ &\int{\sec^2{ax} \,\diff{x}} &&= && \frac{1}{a} \tan{ax} \\ &\int{\tan{ax} \,\diff{x}} &&= && \frac{1}{a} \ln{\abs{\sec{ax}}} \end{alignat*}

coshaxdx=1asinhaxsinhaxdx=1acoshaxsech2axdx=1atanhaxtanhaxdx=1aln(coshax)\begin{alignat*}{2} &\int{\cosh{ax} \,\diff{x}} &&= \frac{1}{a} \sinh{ax} \\ &\int{\sinh{ax} \,\diff{x}} &&= \frac{1}{a} \cosh{ax} \\ &\int{\sech^2{ax} \,\diff{x}} &&= \frac{1}{a} \tanh{ax} \\ &\int{\tanh{ax} \,\diff{x}} &&= \frac{1}{a} \ln{\parens{\cosh{ax}}} \end{alignat*}

cscaxcotaxdx=1acscaxsecaxtanaxdx=1asecaxcsc2axdx=1acotax\begin{alignat*}{3} &\int{\csc{ax} \cot{ax} \,\diff{x}} &&= -&& \frac{1}{a} \csc{ax} \\ &\int{\sec{ax} \tan{ax} \,\diff{x}} &&= && \frac{1}{a} \sec{ax} \\ &\int{\csc^2{ax} \,\diff{x}} &&= -&& \frac{1}{a} \cot{ax} \end{alignat*}

cschaxcothaxdx=1acschaxsechaxtanhaxdx=1asechaxcsch2axdx=1acothax\begin{alignat*}{3} &\int{\csch{ax} \coth{ax} \,\diff{x}} &&= -&& \frac{1}{a} \csch{ax} \\ &\int{\sech{ax} \tanh{ax} \,\diff{x}} &&= -&& \frac{1}{a} \sech{ax} \\ &\int{\csch^2{ax} \,\diff{x}} &&= -&& \frac{1}{a} \coth{ax} \end{alignat*}

secaxdx=1alnsecax+tanaxcscaxdx=1alncotax2=1alncscax+cotaxcotaxdx=1alnsinax\begin{alignat*}{3} &\int{\sec{ax} \,\diff{x}} &&= && \frac{1}{a} \ln{\abs{\sec{ax} + \tan{ax}}} \\ &\int{\csc{ax} \,\diff{x}} &&= -&& \frac{1}{a} \ln{\abs{\cot{\frac{ax}{2}}}} \\ & &&= -&& \frac{1}{a} \ln{\abs{\csc{ax} + \cot{ax}}} \\ &\int{\cot{ax} \,\diff{x}} &&= && \frac{1}{a} \ln{\abs{\sin{ax}}} \end{alignat*}

sechaxdx=1atan1(sinhax)=2atan1(tanhax2)cschaxdx=1alntanhax2=1alncschaxcothaxcothaxdx=1alnsinhax\begin{alignat*}{3} &\int{\sech{ax} \,\diff{x}} &&= \frac{1}{a} \tan^{-1}{\parens{\sinh{ax}}} \\ & &&= \frac{2}{a} \tan^{-1}{\parens{\tanh{\frac{ax}{2}}}} \\ &\int{\csch{ax} \,\diff{x}} &&= \frac{1}{a} \ln{\abs{\tanh{\frac{ax}{2}}}} \\ & &&= \frac{1}{a} \ln{\abs{\csch{ax} - \coth{ax}}} \\ &\int{\coth{ax} \,\diff{x}} &&= \frac{1}{a} \ln{\abs{\sinh{ax}}} \end{alignat*}

1x2+a2dx=1atan1xa1a2x2dx=sin1xa+C1,a>0, a<x<a=cos1xa+C2,a>0, a<x<a1a2x2dx=1atanh1xa+C1,a>x=1acoth1xa+C2,x>a>0=12alna+xax+C3,x2a21x2+a2dx=sinh1xa+C1=ln(x+x2+a2)+C21x2a2dx=cosh1xa+C1,x>a>0=ln(x+x2a2)+C2,x>a>0\begin{alignat*}{3} &\int{\frac{1}{x^2 + a^2} \,\diff{x}} &&= \frac{1}{a} \tan^{-1}{\frac{x}{a}} && \\[\TmpGap] %\vphantom{ % = \frac{1}{a} \coth^{-1}{\frac{x}{a}} + C_2 %} \\ %\vphantom{ % = \frac{1}{2a} \ln{\abs{\frac{a+x}{a-x}}} + C_3 %} \\ &\int{\frac{1}{\sqrt{a^2 - x^2}} \,\diff{x}} &&= \hphantom{-} \sin^{-1}{\frac{x}{a}} + C_1 ,\quad &&\Exn{ a > 0 ,\ -a < x < a } \\ & &&= - \cos^{-1}{\frac{x}{a}} + C_2 ,\quad &&\Exn{ a > 0 ,\ -a < x < a } \\[\TmpGap] &\int{\frac{1}{a^2 - x^2} \,\diff{x}} &&= \frac{1}{a} \tanh^{-1}{\frac{x}{a}} + C_1 ,\quad &&\Exn{ a > \abs{x} } \\ & &&= \frac{1}{a} \coth^{-1}{\frac{x}{a}} + C_2 ,\quad &&\Exn{ \abs{x} > a > 0 } \\ & &&= \frac{1}{2a} \ln{\abs{\frac{a+x}{a-x}}} + C_3 ,\quad &&\Exn{ x^2 \neq a^2 } \\[\TmpGap] &\int{\frac{1}{\sqrt{x^2 + a^2}} \,\diff{x}} &&= \sinh^{-1}{\frac{x}{a}} + C_1 && \\ & &&= \ln{\parens{x + \sqrt{x^2 + a^2}}} + C_2 && \\[\TmpGap] &\int{\frac{1}{\sqrt{x^2 - a^2}} \,\diff{x}} &&= \cosh^{-1}{\frac{x}{a}} + C_1 ,\quad &&\Exn{ x > a > 0 } \\ & &&= \ln{\parens{x + \sqrt{x^2 - a^2}}} + C_2 ,\quad &&\Exn{ x > a > 0 } \end{alignat*}

First Principles

df(x)dx=limh0f(x+h)f(x)h=limuxf(u)f(x)ux \frac{\diff{f\parens{x}}}{\diff{x}} %= \lim_{\delta x \to 0}{\frac{\delta y}{\delta x}} = \lim_{h \to 0}{\frac{f(x+h) - f(x)}{h}} = \lim_{u \to x}{\frac{f(u) - f(x)}{u-x}}

Chain Rule

dydx=dydu×dudx \frac{\memphRC{\diff{y}}}{\memphG{\diff{x}}} = \frac{\memphRC{\diff{y}}}{\memphBC{\diff{u}}} \times \frac{\memphBC{\diff{u}}}{\memphG{\diff{x}}}

Product Rule

ddx(uv)=udvdx+vdudxddx(uvw)=vwdudx+uwdvdx+uvdwdx\begin{gather*} \frac{\diff{}}{\diff{x}} \parens{\memphRC{u} \memphG{v}} = \memphRC{u} \memphG{ \frac{\diff{v}}{\diff{x}}} + \memphG{v} \memphRC{\frac{\diff{u}}{\diff{x}}} \\ \frac{\diff{}}{\diff{x}} \parens{\memphRC{u} \memphG{v} \memphBC{w}} = \memphG{v} \memphBC{w} \memphRC{\frac{\diff{u}}{\diff{x}}} + \memphRC{u} \memphBC{w} \memphG{ \frac{\diff{v}}{\diff{x}}} + \memphRC{u} \memphG{v} \memphBC{\frac{\diff{w}}{\diff{x}}} \end{gather*}

Quotient Rule

ddx(au)=udadxadudxu2=uaauu2 \frac{\diff{}}{\diff{x}}{\parens{\frac{\memphRC{a}}{\memphBC{u}}}} = \frac{ \displaystyle \memphBC{u} \memphRC{\frac{\diff{a}}{\diff{x}}} - \memphRC{a} \memphBC{\frac{\diff{u}}{\diff{x}}} }{ \memphBC{u}^2 } = \frac{ \memphBC{u} \memphRC{a'} - \memphRC{a} \memphBC{u'} }{ \memphBC{u}^2 }

Notation of Second Order Partial Derivatives

Consider a function of two variables f(x,y)f(x, y).

2fx2=x(fx)=(fx)x=fxx2fy2=y(fy)=(fy)y=fyy2fyx=y(fx)=(fx)y=fxy2fxy=x(fy)=(fy)x=fyx\begin{alignat*}{4} \frac{\DsqFun}{\DXsq} &= \frac{\partial{}}{\DX} && \parens{\frac{\DFun}{\DX}} & &= (\memphG{f}_{\X})_{\X} & &= \memphG{f}_{\X \X} \\ \frac{\DsqFun}{\DYsq} &= \frac{\partial{}}{\DY} && \parens{\frac{\DFun}{\DY}} & &= (\memphG{f}_{\Y})_{\Y} & &= \memphG{f}_{\Y \Y} \\[2em] \frac{\DsqFun}{\DY \, \DX} &= \frac{\partial{}}{\DY} && \parens{\frac{\DFun}{\DX}} & &= (\memphG{f}_{\X})_{\Y} & &= \memphG{f}_{\X \Y} \\ \frac{\DsqFun}{\DX \, \DY} &= \frac{\partial{}}{\DX} && \parens{\frac{\DFun}{\DY}} & &= (\memphG{f}_{\Y})_{\X} & &= \memphG{f}_{\Y \X} \end{alignat*}

Equality of Mixed Partials

Consider a function of two variables f(x,y)f(x, y).

If ff and all of its first and second order partial derivatives are continuous, then:

2fxy=2fyx \frac{\DsqFun}{\DX \, \DY} = \frac{\DsqFun}{\DY \, \DX}

TODO: This apparently generalizes to higher derivatives. Rewrite this and find a good source about continuity requirements.

Reverse Chain Rule

f(x)g(f(x))dx=g(f(x))+C \int{\memphRC{f'(x)} \, \memphBC{g(}\memphRC{f(x)}\memphBC{)} \,\diff{x}} = \memphBC{g'(}\memphRC{f(x)}\memphBC{)} + C

Integration By Substitution

u:=φ(x)abf(x)dx=abφ(x)f(φ(x))dx=abdudxf(u)dx=φ(a)φ(b)f(u)du \memphRC{u} \defeq \memphRC{\varphi(x)} \\ \begin{aligned} \int_a^b{\memphBC{f(x)} \,\diff{x}} &= \int_a^b{ \memphRC{\varphi'(x)} \, \memphBC{f(\memphRC{\varphi(x)})} \, \diff{x} } \\ &= \int_a^b{ \memphRC{\frac{\diff{u}}{\diff{x}}} \, \memphBC{f(\memphRC{u})} \, \diff{x} } = \int_{\memphRC{\varphi(a)}}^{\memphRC{\varphi(b)}}{ \, \memphBC{f(\memphRC{u})} \, \memphRC{\diff{u}} } \end{aligned}

Integration By Parts

f(x)G(x)dx=F(x)G(x)F(x)g(x)dx \int{\memphRC{f(x)} \, \memphBC{G(x)} \,\diff{x}} = \memphRC{F(x)} \, \memphBC{G(x)} - \int{\memphRC{F(x)} \, \memphBC{g(x)} \,\diff{x}}

Flipped Integration Bounds

abf(x)dx=baf(x)dx \int_a^b f\parens{x} \,\diff{x} = - \int_b^a f\parens{x} \,\diff{x}

Leibniz Integral Rule

ddxb(x)a(x)f(x,y)dy=f(x,a(x))ddxa(x)f(x,b(x))ddxb(x)+b(x)a(x)xf(x,y)dy\begin{align*} &\frac{\diff{}}{\diff{x}} \int_{b\parens{x}}^{a\parens{x}} f\parens{x,y} \,\diff{y} = f \parens{x, a\parens{x}} \frac{\diff{}}{\diff{x}} a\parens{x} \\ &\qquad - f \parens{x, b\parens{x}} \frac{\diff{}}{\diff{x}} b\parens{x} + \int_{b\parens{x}}^{a\parens{x}}{ \frac{\partial{}}{\partial{x}} f\parens{x,y} \,\diff{y} } \end{align*}

If the limits are constants:

ddxbaf(x,y)dy=baxf(x,y)dy\begin{gather*} \frac{\diff{}}{\diff{x}} \int_b^a f\parens{x,y} \,\diff{y} = \int_b^a \frac{\partial{}}{\partial{x}} f\parens{x,y} \,\diff{y} %\\ %\frac{\diff{}}{\diff{x}} %\int_b^x f\parens{x,y} \,\diff{y} %= f\parens{x, x} %+ \int_b^x \frac{\partial{}}{\partial{x}} f\parens{x,y} \,\diff{y} \end{gather*}

Separable ODEs

Write the ODE in the form:

h(y)dy=g(x)dx h(y) \,\diff{y} = g(x) \,\diff{x}

To solve, integrate both sides:

H(y)=G(x)+C H(y) = G(x) + C

First-Order Linear ODEs

Write the ODE in the form:

dydx+f(x)y=g(x) \frac{\diff{y}}{\diff{x}} + f(x) \, y = g(x)

First, define the integrating factor hh:

h(x)=ef(x)dxh(x)=f(x)h(x) h(x) = e^{\int{f(x) \,\diff{x}}} \Exn{ \qquad \Longrightarrow \qquad h'(x) = f(x) \, h(x) }

Multiply the ODE by hh and undo the product rule:

h(x)dydx+h(x)f(x)y=g(x)h(x)ddx(h(x)y)=g(x)h(x) h(x) \, \frac{\diff{y}}{\diff{x}} + h(x) \, f(x) \, y = g(x) \, h(x) \\ \frac{\diff{}}{\diff{x}} \parens{h(x) \, y} = g(x) \, h(x)

To solve, integrate both sides:

h(x)y=g(x)h(x)dx h(x) \, y = \int{g(x) \, h(x) \,\diff{x}}

Exact ODEs

Write the ODE in the form:

F(x,y)+G(x,y)dydx=0 F(x, y) + G(x, y) \frac{\diff{y}}{\diff{x}} = 0

First prove the ODE is exact by showing:

Fy=Gx=2Hxy=2Hyxequality ofmixed partials \frac{\partial{F}}{\partial{y}} = \frac{\partial{G}}{\partial{x}} \Exn{ {} = \underbrace{ \frac{\partial^2{H}}{\partial{x} \, \partial{y}} = \frac{\partial^2{H}}{\partial{y} \, \partial{x}} }_{\substack{\text{equality of}\\\text{mixed partials}}} }

Thus, our ODE can be rewritten:

Hx(x,y)+Hy(x,y)dydx=0 H_x(x, y) + H_y(x, y) \frac{\diff{y}}{\diff{x}} = 0

This can be simplified by undoing the chain rule:

Hxdxdx+Hydydx=0ddx(H(x,y(x)))=0\begin{gather*} \frac{\partial{H}}{\partial{x}} \frac{\diff{x}}{\diff{x}} + \frac{\partial{H}}{\partial{y}} \frac{\diff{y}}{\diff{x}} = 0 \\ \frac{\diff{}}{\diff{x}} \parens{H \parens{x, y(x)}} = 0 \end{gather*}

Therefore, the solution is:

H(x,y)=C H(x, y) = C

such that:

H(x,y)=Hx(x,y)x=H1(x,y)+C1(y)H(x,y)=Hy(x,y)y=H2(x,y)+C2(x)\begin{alignat*}{3} H(x, y) &= \int{H_x(x, y) \,\partial{x}} &&= H_1(x, y) &&+ C_1(y) \\ H(x, y) &= \int{H_y(x, y) \,\partial{y}} &&= H_2(x, y) &&+ C_2(x) \end{alignat*}

Trigonometric Fourier Series

For any periodic function f(t)=f(t+nT)f \parens{t} = f \parens{t + n T}:

f(t)=ao+n=1(ancosnω0t+bnsinnω0t)=ao+n=1Ancos(nω0t+ϕn)(Amplitude-Phase Form)\begin{align*} f \parens{t} &= a_o + \sum_{n=1}^\infty \parens{ a_n \cos n \omega_0 t + b_n \sin n \omega_0 t } \\ &= a_o + \sum_{n=1}^\infty { A_n \cos \parens{n \omega_0 t + \phi_n} } \qquad\quad \begin{array}{l} \text{\footnotesize{}(Amplitude-} \\ \text{\footnotesize{}Phase Form)} \\ \end{array} %\htmlStyle{width: 5em;}{\text{\footnotesize{}(Amplitude-Phase Form)}} \end{align*}

a0=1T0Tf(t)dt=average value of fan=2T0Tf(t)cosnω0tdtbn=2T0Tf(t)sinnω0tdtAn=an2+bn2ϕn=tan1bnan\begin{gather*} a_0 = \frac{1}{T} \int_0^T f\parens{t} \,\diff{t} = \text{average value of $f$} \\ a_n = \frac{2}{T} \int_0^T f\parens{t} \cos n \omega_0 t \,\diff{t} \\ b_n = \frac{2}{T} \int_0^T f\parens{t} \sin n \omega_0 t \,\diff{t} \\ A_n = \sqrt{a_n^2 + b_n^2} \qquad\quad \phi_n = - \tan^{-1} \frac{b_n}{a_n} \end{gather*}

Fundamental angular frequency:

ω0=2πT \omega_0 = \frac{2 \pi}{T}

Each element of the sum is called a harmonic.

Dirichlet conditions (for a convergent Fourier series):

  • ff must be single-valued everywhere.
  • ff must have a finite number of finite discontinuities, maxima, and minima in any one period.
  • t0t0+Tf(t)dt<\int_{t_0}^{t_0+T} \abs{f\parens{t}} \,\diff{t} < \infty for any t0t_0.

Exponential Fourier Series

Alternative to the trigonometric Fourier series:

f(t)=n=cneinω0t=c0+n=1(cneinω0t+cneinω0t) f\parens{t} = \sum_{n = -\infty}^\infty c_n e^{i n \omega_0 t} = c_0 + \sum_{n = 1}^\infty \parens{ c_n e^{i n \omega_0 t} + c_{-n} e^{i n \omega_0 t} }

c0=a0cn=1T0Tf(t)einω0tdt,n0 c_0 = a_0 \qquad\quad c_n = \frac{1}{T} \int_0^T f\parens{t} \, e^{-i n \omega_0 t} \,\diff{t} , \quad \forall n \ne 0

Related by:

Anϕn=anibn=2cn,n0 A_n \phase{\phi_n} = a_n - i b_n = 2 c_n, \quad \forall n \ne 0

Fourier Series: Useful Integral Identities

n,mZ0Tsinnω0tdt=00Tcosnω0tdt=00Tsinnω0tcosmω0tdt=00Tsinnω0tsinmω0tdt=0,(nm)0Tcosnω0tcosmω0tdt=0,(nm)0Tsin2nω0tdt=T20Tcos2nω0tdt=T2tcosatdt=1a2cosat+1atsinat+Ctsinatdt=1a2sinat1atcosat+C n, m \in \myIntegerSet \\[0.4em] \int_0^T \sin{n \omega_0 t} \,\diff{t} = 0 \\ \int_0^T \cos{n \omega_0 t} \,\diff{t} = 0 \\ \int_0^T \sin{n \omega_0 t} \, \cos{m \omega_0 t} \,\diff{t} = 0 \\ \int_0^T \sin{n \omega_0 t} \, \sin{m \omega_0 t} \,\diff{t} = 0 , \qquad \parens{n \ne m} \\ \int_0^T \cos{n \omega_0 t} \, \cos{m \omega_0 t} \,\diff{t} = 0 , \qquad \parens{n \ne m} \\ \int_0^T \sin^2{n \omega_0 t} \,\diff{t} = \frac{T}{2} \\ \int_0^T \cos^2{n \omega_0 t} \,\diff{t} = \frac{T}{2} \\[1.8em] \int t \cos{a t} \,\diff{t} = \frac{1}{a^2} \cos{a t} + \frac{1}{a} t \sin{a t} + C \\ \int t \sin{a t} \,\diff{t} = \frac{1}{a^2} \sin{a t} - \frac{1}{a} t \cos{a t} + C

Fourier Transform

For function ff (not necessarily periodic) which we interpret as a periodic function with infinite period:

F(ω)=F[f(t)]=f(t)eiωtdtf(t)=F1[F(ω)]=12πF(ω)eiωtdω\begin{gather*} F\parens{\omega} = \FourierTransform\brackets{f\parens{t}} = \int_{-\infty}^\infty f\parens{t} \, e^{-i \omega t} \,\diff{t} \\ f\parens{t} = \FourierTransform^{-1}\brackets{F\parens{\omega}} = \frac{1}{2 \pi} \int_{-\infty}^\infty F\parens{\omega} \, e^{i \omega t} \,\diff{\omega} \end{gather*}

The Fourier transform FF exists when the Fourier integral converges.

Heaviside Step Function

H(x):={1if x>00if x0 H\parens{x} \defeq \begin{cases} 1 &\quad \text{if } x > 0 \\ 0 &\quad \text{if } x \le 0 \end{cases}

Commonly also written as u(x)u\parens{x} or θ(x)\theta\parens{x}.

Dirac Delta Function

δ(x)=ddxH(x) \delta\parens{x} = \frac{\diff{}}{\diff{x}} H\parens{x}

Sign/Signum Function

sgn(x):=u(t)u(t)={1if x<00if x=01if x>0 \SignFunction\parens{x} \defeq u\parens{t} - u\parens{-t} = \begin{cases} -1 &\quad \text{if } x < 0 \\ 0 &\quad \text{if } x = 0 \\ 1 &\quad \text{if } x > 0 \end{cases}

Convolution Integral

h(t)x(t)=x(λ)h(tλ)dλ h\parens{t} * x\parens{t} = \int_{-\infty}^\infty{ x\parens{\lambda} \, h\parens{t - \lambda} \,\diff{\lambda} } %% TODO: I found this simplification in a textbook, but %% I'm not sure if it's right. %= \underbrace{ % \vphantom{\frac{1}{\displaystyle \frac{1}{1}}} % \int_0^t{ % x\parens{\lambda} h\parens{t - \lambda} \,\diff{\lambda} % } %}_{\text{if $h\parens{t} = 0$ for $t < 0$}}

Useful properties:

xh=hxf[x+y]=fx+fyf[xy]=[fx]y x * h = h * x \\ f * \brackets{x + y} = f * x + f * y \\ f * \brackets{x * y} = \brackets{f * x} * y

f(t)δ(tt0)=f(tt0)f(t)δ(t)=f(λ)δ(tλ)dλ=f(t)f(t)u(t)=f(λ)u(tλ)dλ=tf(λ)dλ f\parens{t} * \delta\parens{t - t_0} = f\parens{t - t_0} \\ f\parens{t} * \delta'\parens{t} = \int_{-\infty}^\infty { f\parens{\lambda} \, \delta'\parens{t - \lambda} \,\diff{\lambda} } = f'\parens{t} \\ f\parens{t} * u\parens{t} = \int_{-\infty}^\infty { f\parens{\lambda} \, u\parens{t - \lambda} \,\diff{\lambda} } = \int_{-\infty}^t f\parens{\lambda} \,\diff{\lambda}

Fourier Transform Properties

F(t)F2πf(ω)Dualityaf(t)+bg(t)FaF(ω)+bG(ω)Linearityf(at)F1aF(ωa)TimeScalingf(tt0)Feiωt0F(ω)TimeShiftingf(t)eiω0tFF(ωω0)AmplitudeModulation(FrequencyShifting)f(n)(t)F(iω)nF(ω)TimeDifferentiationtf(τ)dτFF(ω)iω+πF(0)δ(ω)TimeIntegrationf(t)FF(ω)=F(ω)Reversalf(t)g(t)FF(ω)G(ω)ConvolutionIn tf(t)g(t)F12πF(ω)G(ω)ConvolutionIn ω \begin{array}{cccl} F\parens{t} &\TmpFTArrow& 2 \pi f \parens{-\omega} & \begin{array}{l} \text{\tiny\myul{Duality}} \end{array} \\[0.6em] a \, f(t) + b \, g(t) &\TmpFTArrow& a \, F\parens{\omega} + b \, G\parens{\omega} & \begin{array}{l} \text{\tiny{Linearity}} \end{array} \\[0.6em] {\displaystyle f\parens{a t}} &\TmpFTArrow& {\displaystyle \frac{1}{\abs{a}} F\parens{\frac{\omega}{a}}} & \begin{array}{l} \text{\tiny{Time}} \\[\TmpTextBreak] \text{\tiny{Scaling}} \end{array} \\[0.6em] f\parens{t - t_0} &\TmpFTArrow& e^{-i \omega t_0} F\parens{\omega} & \begin{array}{l} \text{\tiny{Time}} \\[\TmpTextBreak] \text{\tiny{Shifting}} \end{array} \\[0.6em] f\parens{t} e^{i \omega_0 t} &\TmpFTArrow& F\parens{\omega - \omega_0} & \begin{array}{l} \text{\tiny{Amplitude}} \\[\TmpTextBreak] \text{\tiny{Modulation}} \\[\TmpTextBreak] \text{\tiny{(Frequency}} \\[\TmpTextBreak] \text{\tiny{Shifting)}} \end{array} \\[0.6em] f^{(n)}\parens{t} &\TmpFTArrow& \parens{i \omega}^n F\parens{\omega} & \begin{array}{l} \text{\tiny{Time}} \\[\TmpTextBreak] \text{\tiny{Differentiation}} \end{array} \\[0.6em] {\displaystyle \int_{-\infty}^t f\parens{\tau} \,\diff{\tau}} &\TmpFTArrow& {\displaystyle \frac{F\parens{\omega}}{i \omega} + \pi F\parens{0} \, \delta\parens{\omega} } & \begin{array}{l} \text{\tiny{Time}} \\[\TmpTextBreak] \text{\tiny{Integration}} \end{array} \\[0.6em] f\parens{-t} &\TmpFTArrow& F\parens{-\omega} = F^*\parens{-\omega} & \begin{array}{l} \text{\tiny{Reversal}} \end{array} \\[0.6em] f\parens{t} * g\parens{t} &\TmpFTArrow& F\parens{\omega} \, G\parens{\omega} & \begin{array}{l} \text{\tiny{Convolution}} \\[\TmpTextBreak] \text{\tiny{In $t$}} \end{array} \\[0.6em] {\displaystyle f\parens{t} \, g\parens{t}} &\TmpFTArrow& {\displaystyle \frac{1}{2 \pi} F\parens{\omega} * G\parens{\omega} } & \begin{array}{l} \text{\tiny{Convolution}} \\[\TmpTextBreak] \text{\tiny{In $\omega$}} \end{array} \end{array}

Fourier Transform Common Pairs

δ(t)F1F2πδ(ω)u(t)Fπδ(ω)+1jωu(t+τ)u(tτ)F2sinωτωtF2ω2sgn(t)F2iωeatu(t)F1a+iωeatu(t)F1aiωtneatu(t)Fn!(a+iω)n+1eatF2aa2+ω2eiω0tF2πδ(ωω0)sinω0tFiπ[δ(ω+ω0)δ(ωω0)]cosω0tFπ[δ(ω+ω0)+δ(ωω0)]eatsin(ω0t)u(t)Fω0(a+iω)2+ω02eatcos(ω0t)u(t)Fa+iω(a+iω)2+ω02 \begin{array}{ccc} \delta\parens{t} &\TmpFTArrow& 1 \qquad\TmpFTArrow\qquad 2 \pi \, \delta\parens{\omega} \\[1em] u\parens{t} &\TmpFTArrow& \displaystyle \pi \, \delta\parens{\omega} + \frac{1}{j \omega} \\[1em] u\parens{t + \tau} - u\parens{t - \tau} &\TmpFTArrow& \displaystyle 2 \frac{\sin{\omega \tau}}{\omega} \\[1em] \abs{t} &\TmpFTArrow& \displaystyle \frac{- 2}{\omega^2} \\[1em] \SignFunction\parens{t} &\TmpFTArrow& \displaystyle \frac{2}{i \omega} \\[1em] e^{-a t} \, u\parens{t} &\TmpFTArrow& \displaystyle \frac{1}{a + i \omega} \\[1em] e^{a t} \, u\parens{-t} &\TmpFTArrow& \displaystyle \frac{1}{a - i \omega} \\[1em] t^n e^{-a t} \, u\parens{t} &\TmpFTArrow& \displaystyle \frac{n!}{\parens{a + i \omega}^{n + 1}} \\[1em] e^{-a \abs{t}} &\TmpFTArrow& \displaystyle \frac{2 a}{a^2 + \omega^2} \\[1em] e^{i \omega_0 t} &\TmpFTArrow& 2 \pi \, \delta\parens{\omega - \omega_0} \\[1em] \sin{\omega_0 t} &\TmpFTArrow& i \pi \brackets{ \delta\parens{\omega + \omega_0} - \delta\parens{\omega - \omega_0} } \\[1em] \cos{\omega_0 t} &\TmpFTArrow& \pi \brackets{ \delta\parens{\omega + \omega_0} + \delta\parens{\omega - \omega_0} } \\[1em] e^{-a t} \sin\parens{\omega_0 t} \, u\parens{t} &\TmpFTArrow& \displaystyle \frac{\omega_0}{\parens{a + i \omega}^2 + \omega_0^2} \\[1em] e^{-a t} \cos\parens{\omega_0 t} \, u\parens{t} &\TmpFTArrow& \displaystyle \frac{a + i \omega}{\parens{a + i \omega}^2 + \omega_0^2} \end{array}

Discrete Fourier Transform (DFT)

Instead of a continuous function over time, suppose we instead discretely sample it, denoted xnx_n. The DFT of xnx_n is:

Xn=k=0N1xkei2πNnk=k=0N1xk[cos(2πNkn)isin(2πNkn)]\begin{align*} X_n &= \sum_{k=0}^{N-1} x_k e^{-i \frac{2 \pi}{N} n k} \\ &= \sum_{k=0}^{N-1} x_k \brackets{ \cos\parens{\frac{2 \pi}{N} k n} - i \sin\parens{\frac{2 \pi}{N} k n} } \end{align*}

Side-note: Fast Fourier Transform (FFT) is a family of algorithms used to calculate the DFT.

TODO: What are some better definitions? Can values be complex?

Laplace Transform

For function ff and some complex variable s=σ+iωs = \sigma + i \omega:

L[f(t)]=F(s)=0f(t)estdt \LaplaceTransform\brackets{f\parens{t}} = F\parens{s} = \int_{0^-}^\infty f\parens{t} e^{-s t} \,\diff{t}

Laplace Transform Properties

Linearity:

L[a1f1(t)+a2f2(t)]=a1F1(s)+a2F2(s) \LaplaceTransform\brackets{a_1 f_1(t) + a_2 f_2(t)} = a_1 F_1\parens{s} + a_2 F_2\parens{s}

Scaling:

L[f(at)]=1aF(sa) \LaplaceTransform\brackets{f\parens{a t}} = \frac{1}{a} F\parens{\frac{s}{a}}

Time Shifting:

L[f(ta)u(ta)]=easF(s) \LaplaceTransform\brackets{f\parens{t - a} \, u\parens{t - a}} = e^{-a s} F\parens{s}

Frequency Shifting/Translation:

L[eatf(t)u(t)]=F(s+a) \LaplaceTransform\brackets{e^{-a t} f\parens{t} \, u\parens{t}} = F\parens{s + a}

Time Differentiation:

L[f(t)]=sF(s)f(0)L[f(t)]=s2F(s)sf(0)f(0)\begin{gather*} \LaplaceTransform\brackets{f'\parens{t}} = s F\parens{s} - f\parens{0^-} \\ \LaplaceTransform\brackets{f''\parens{t}} = s^2 F\parens{s} - s f\parens{0^-} - f'\parens{0^-} \end{gather*}

L[f(n)(t)]=snF(s)sn1f(0)sn2f(0)=s0f(n1)(0)\begin{align*} \LaplaceTransform\brackets{f^{\parens{n}}\parens{t}} &= s^n F\parens{s} - s^{n-1} f\parens{0^-} - s^{n-2} f'\parens{0^-} \\ &\phantom{=} - \cdots - s^0 f^{\parens{n-1}}\parens{0^-} \end{align*}

Time Integration:

L[0tf(x)dx]=1sF(s) \LaplaceTransform\brackets{\int_0^t f\parens{x} \,\diff{x}} = \frac{1}{s} F\parens{s}

Frequency Differentiation:

L[tnf(t)]=(1)nF(n)(s) \LaplaceTransform\brackets{t^n f\parens{t}} = \parens{-1}^n F^{\parens{n}}\parens{s}

Time Periodicity:

TODO: Write better notes on Time Periodicity?

F(s)=F1(s)1eTs F\parens{s} = \frac{F_1\parens{s}}{1 - e^{-T s}}

Initial/Final Values:

f(0)=limssF(s)f()=lims0sF(s)\begin{gather*} f\parens{0} = \lim_{s \to \infty} s F\parens{s} \qquad\quad f\parens{\infty} = \lim_{s \to 0} s F\parens{s} \end{gather*}

Convolution:

TODO: Write better notes and properties of the convolution?

F1(s)F2(s)=L[f1(t)f2(t)] F_1\parens{s} \, F_2\parens{s} = \LaplaceTransform\brackets{f_1\parens{t} * f_2\parens{t}}