Proof #1

L{1}=∫0∞(1⋅e−st)dt=[1−se−st]0∞=0−(1−s)=1sL\left\{1\right\}=\int _0^{\infty }\left(1\cdot e^{-st}\right)dt\:=\left[\frac{1}{-s}e^{-st}\right]_0^{\infty }=0-\left(\frac{1}{-s}\right)=\frac{1}{s}

Proof #2

L{t}=∫0∞(t⋅e−st)dtL\left\{t\right\}=\int _0^{\infty \:}\left(t\cdot \:e^{-st}\right)dt u=tu=t dv=e−stdv=e^{-st} du=dtdu=dt v=1−se−stv=\frac{1}{-s}e^{-st} L{t}=[t⋅1−se−st]0∞−∫(1−se−st)dtL\left\{t\right\}=\left[t\cdot \:\:\:\frac{1}{-s}e^{-st}\right]^{\infty \:}_0-\int \:\:\left(\frac{1}{-s}e^{-st}\right)dt L{t}=(0−0)−(1−s)∫(e−st)dtL\left\{t\right\}=\left(0-0\right)-\left(\frac{1}{-s}\right)\int \:\:\:\left(e^{-st}\right)dt L{t}=(1s)∫(1⋅e−st)dt=1sL{1}L\left\{t\right\}=\left(\frac{1}{s}\right)\int \:\:\:\:\:\left(1\cdot \:\:e^{-st}\right)dt=\frac{1}{s}L\left\{1\right\} L{t}=1s⋅1s=1s2L\left\{t\right\}=\frac{1}{s}\cdot \frac{1}{s}=\frac{1}{s^2}

Proof #3

L{e−αt}=∫0∞(e−αte−st)dt=∫0∞(e(−α−s)t)dtL\left\{e^{-\alpha t}\right\}=\int _0^{\infty }\left(e^{-\alpha t}e^{-st}\right)\:dt=\int _0^{\infty \:}\left(e^{\left(-\alpha \:-s\right)t}\right)\:dt L{e−αt}=[1−α−se(−α−s)t]0∞=(0−(1−α−s))L\left\{e^{-\alpha \:t}\right\}=\left[\frac{1}{-\alpha \:-s}e^{\left(-\alpha \:\:\:\:-s\right)t}\right]_0^{\infty \:}=\left(0-\left(\frac{1}{-\alpha \:-s}\right)\right) L{e−αt}=1α+sL\left\{e^{-\alpha \:t}\right\}=\frac{1}{\alpha +s}

Proof #4

L{sin(αt)}=∫0∞(sin(αt)⋅e−st)dtL\left\{sin\left(\alpha \:t\right)\right\}=\int _0^{\infty \:}\left(sin\left(\alpha \:\:t\right)\cdot \:\:e^{-st}\right)dt u=e−stu=e^{-st} dv=sin(αt)dtdv=\:sin\left(\alpha t\right)dt du=−se−stdtdu=-se^{-st}dt v=−cos(αt)αv=\frac{-cos\left(\alpha \:t\right)}{\alpha \:} ∫(sin(αt)e−st)dt=(e−st⋅(−cos(αt)α))−(∫−cos(αt)α⋅(−se−st)dt)\int \left(sin\left(\alpha t\right)e^{-st}\right)dt=\left(e^{-st}\cdot \left(\frac{-cos\left(\alpha t\right)}{\alpha }\right)\right)-\left(\int \:\frac{-cos\left(\alpha t\right)}{\alpha }\cdot \left(-se^{-st}\right)dt\right) ∫(sin(αt)e−st)dt=(−e−stcos(αt)α)−(sa∫cos(αt)e−stdt)\int \left(sin\left(\alpha t\right)e^{-st}\right)dt=\left(\frac{-e^{-st}cos\left(\alpha t\right)}{\alpha }\right)-\left(\frac{s}{a}\int \:\:cos\left(\alpha \:t\right)e^{-st}dt\right) u=e−stu=e^{-st} dv=cos(αt)dtdv=\:cos\left(\alpha t\right)dt du=−se−stdtdu=-se^{-st}dt v=sin(αt)αv=\frac{sin\left(\alpha t\right)}{\alpha } ∫(sin(αt)e−st)dt=(−e−stcos(αt)α)−(sa(e−st⋅sin(αt)α−∫sin(αt)α(−se−st)dt))\int \:\left(sin\left(\alpha \:t\right)e^{-st}\right)dt=\left(\frac{-e^{-st}cos\left(\alpha \:t\right)}{\alpha \:}\right)-\left(\frac{s}{a}\left(e^{-st}\cdot \frac{sin\left(\alpha \:t\right)}{\alpha \:}-\int \:\frac{sin\left(\alpha \:t\right)}{\alpha \:}\left(-se^{-st}\right)dt\right)\right) L{te−αt}=[te−(α+s)t−(α+s)]0∞+1(α+s)∫0∞e(−α−s)tdtL\left\{te^{-\alpha t}\right\}=\left[\frac{te^{-\left(\alpha +s\right)t}}{-\left(\alpha +s\right)}\right]^{\infty }_0+\frac{1}{\left(\alpha +s\right)}\int _0^{\infty }e^{\left(-\alpha -s\right)t}\:dt ∫(sin(αt)e−st)dt+s2a2∫sin(αt)(e−st)dt=(−e−stcos(αt)α)−se−stsin(αt)a2\int \:\left(sin\left(\alpha t\right)e^{-st}\right)dt+\frac{s^2}{a^2}\int sin\left(\alpha t\right)\left(e^{-st}\right)dt=\left(\frac{-e^{-st}cos\left(\alpha t\right)}{\alpha \:\:}\right)-\frac{se^{-st}sin\left(\alpha t\right)}{a^2}   ∫(sin(αt)e−st)dt(1+s2a2)=(−e−stcos(αt)α)−se−stsin(αt)a2\int \:\:\left(sin\left(\alpha \:t\right)e^{-st}\right)dt\left(1+\frac{s^2}{a^2}\right)=\left(\frac{-e^{-st}cos\left(\alpha \:t\right)}{\alpha \:\:\:}\right)-\frac{se^{-st}sin\left(\alpha \:t\right)}{a^2} ∫(sin(αt)e−st)dt=((−e−stcos(αt)α)−se−stsin(αt)a2)(α2+s2a2)\int \left(sin\left(\alpha t\right)e^{-st}\right)dt=\frac{\left(\left(\frac{-e^{-st}cos\left(\alpha t\right)}{\alpha }\right)-\frac{se^{-st}sin\left(\alpha t\right)}{a^2}\right)}{\left(\frac{\alpha ^2+s^2}{a^2}\right)} L{sin(αt)}=∫0∞(sin(αt)e−st)dt=[((−e−stcos(αt)α)−se−stsin(αt)a2)(α2+s2a2)]0∞L\left\{sin\left(\alpha \:t\right)\right\}=\int _0^{\infty \:}\left(sin\left(\alpha \:\:t\right)e^{-st}\right)dt=\left[\frac{\left(\left(\frac{-e^{-st}cos\left(\alpha t\right)}{\alpha \:\:\:\:\:\:}\right)-\frac{se^{-st}sin\left(\alpha t\right)}{a^2}\right)}{\left(\frac{\alpha ^2+s^2}{a^2}\right)}\right]_0^{\infty \:} L{sin(αt)}=0−−1αα2+s2a2=αα2+s2L\left\{sin\left(\alpha t\right)\right\}=0-\frac{-\frac{1}{\alpha \:\:}}{\frac{\alpha ^2+s^2}{a^2}}=\frac{\alpha }{\alpha \:\:^2+s^2}  

Proof #5

L{cos(αt)}=∫0∞(cos(αt)⋅e−st)dtL\left\{cos\left(\alpha t\right)\right\}=\int _0^{\infty }\left(cos\left(\alpha t\right)\cdot e^{-st}\right)dt u=e−stu=e^{-st} dv=cos(αt)dtdv=\:cos\left(\alpha t\right)dt du=−se−stdtdu=-se^{-st}dt v=sin(αt)αv=\frac{sin\left(\alpha t\right)}{\alpha } L{cos(αt)}=[e−st⋅sin(αt)α]0∞−∫0∞sin(αt)α(−se−st)dtL\left\{cos\left(\alpha \:t\right)\right\}=\left[e^{-st}\cdot \frac{sin\left(\alpha \:t\right)}{\alpha \:}\right]_0^{\infty }-\int _0^{\infty }\frac{sin\left(\alpha \:t\right)}{\alpha \:\:\:}\left(-se^{-st}\right)dt L{cos(αt)}=[e−stsin(αt)α]0∞+sα∫sin(αt)(e−st)dtL\left\{cos\left(\alpha \:t\right)\right\}=\left[\frac{e^{-st}sin\left(\alpha \:t\right)}{\alpha \:\:\:\:\:}\right]_0^{\infty }+\frac{s}{\alpha }\int \:sin\left(\alpha \:t\right)\left(e^{-st}\right)dt L{cos(αt)}=[e−stsin(αt)α]0∞+sα⋅L{sin(αt)}L\left\{cos\left(\alpha \:t\right)\right\}=\left[\frac{e^{-st}sin\left(\alpha \:\:t\right)}{\alpha \:\:\:\:\:\:}\right]_0^{\infty }+\frac{s}{\alpha \:}\cdot \:L\left\{sin\left(\alpha \:t\right)\right\} L{cos(αt)}=[e−stsin(αt)α]0∞+sα⋅αα2+s2L\left\{cos\left(\alpha \:t\right)\right\}=\left[\frac{e^{-st}sin\left(\alpha \:\:t\right)}{\alpha \:\:\:\:\:\:}\right]_0^{\infty }+\frac{s}{\alpha \:}\cdot \frac{\alpha }{\alpha \:^2+s^2} L{cos(αt)}=(0−0)+sα2+s2=sα2+s2L\left\{cos\left(\alpha \:t\right)\right\}=\left(0-0\right)+\frac{s}{\alpha \:^2+s^2}=\frac{s}{\alpha \:\:^2+s^2}

Proof #6

L{te−αt}=∫0∞te−αte−st=∫0∞te(−α−s)tL\left\{te^{-\alpha t}\right\}=\int _0^{\infty }te^{-\alpha t}e^{-st}=\int _0^{\infty }te^{\left(-\alpha -s\right)t} u=tu=t dv=e(−α−s)tdtdv=e^{\left(-\alpha -s\right)t}dt du=1dtdu=1dt v=1(−α−s)e(−α−s)tv=\frac{1}{\left(-\alpha \:\:-s\right)}e^{\left(-\alpha \:-s\right)t} L{te−αt}=[te−(α+s)t−(α+s)]0∞−∫0∞e−(α+s)t−(α+s)dtL\left\{te^{-\alpha t}\right\}=\left[\frac{te^{-\left(\alpha +s\right)t}}{-\left(\alpha +s\right)}\right]_0^{\infty }-\int _0^{\infty }\frac{e^{-\left(\alpha +s\right)t}}{-\left(\alpha +s\right)}dt L{te−αt}=[te−(α+s)t−(α+s)]0∞+1(α+s)∫0∞e(−α−s)tdtL\left\{te^{-\alpha t}\right\}=\left[\frac{te^{-\left(\alpha +s\right)t}}{-\left(\alpha +s\right)}\right]^{\infty }_0+\frac{1}{\left(\alpha +s\right)}\int _0^{\infty }e^{\left(-\alpha -s\right)t}\:dt L{te−αt}=[te−(α+s)t−(α+s)]0∞+1(α+s)L{e−αt}L\left\{te^{-\alpha t}\right\}=\left[\frac{te^{-\left(\alpha \:\:+s\right)t}}{-\left(\alpha \:\:+s\right)}\right]_0^{\infty \:\:}+\frac{1}{\left(\alpha \:\:+s\right)}L\left\{e^{-\alpha \:\:t}\right\} L{te−αt}=(0−0)+1(α+s)1(α+s)=1(α+s)2L\left\{te^{-\alpha t}\right\}=\left(0-0\right)+\frac{1}{\left(\alpha \:\:\:+s\right)}\frac{1}{\left(\alpha \:\:\:\:+s\right)}=\frac{1}{\left(\alpha \:\:+s\right)^2}

Proof #5

L{cos(αt)}=∫0∞(cos(αt)⋅e−st)dtL\left\{cos\left(\alpha t\right)\right\}=\int _0^{\infty }\left(cos\left(\alpha t\right)\cdot e^{-st}\right)dt u=e−stu=e^{-st} dv=cos(αt)dtdv=\:cos\left(\alpha t\right)dt du=−se−stdtdu=-se^{-st}dt v=sin(αt)αv=\frac{sin\left(\alpha t\right)}{\alpha } L{cos(αt)}=[e−st⋅sin(αt)α]0∞−∫0∞sin(αt)α(−se−st)dtL\left\{cos\left(\alpha \:t\right)\right\}=\left[e^{-st}\cdot \frac{sin\left(\alpha \:t\right)}{\alpha \:}\right]_0^{\infty }-\int _0^{\infty }\frac{sin\left(\alpha \:t\right)}{\alpha \:\:\:}\left(-se^{-st}\right)dt L{cos(αt)}=[e−stsin(αt)α]0∞+sα∫sin(αt)(e−st)dtL\left\{cos\left(\alpha \:t\right)\right\}=\left[\frac{e^{-st}sin\left(\alpha \:t\right)}{\alpha \:\:\:\:\:}\right]_0^{\infty }+\frac{s}{\alpha }\int \:sin\left(\alpha \:t\right)\left(e^{-st}\right)dt L{cos(αt)}=[e−stsin(αt)α]0∞+sα⋅L{sin(αt)}L\left\{cos\left(\alpha \:t\right)\right\}=\left[\frac{e^{-st}sin\left(\alpha \:\:t\right)}{\alpha \:\:\:\:\:\:}\right]_0^{\infty }+\frac{s}{\alpha \:}\cdot \:L\left\{sin\left(\alpha \:t\right)\right\} L{cos(αt)}=[e−stsin(αt)α]0∞+sα⋅αα2+s2L\left\{cos\left(\alpha \:t\right)\right\}=\left[\frac{e^{-st}sin\left(\alpha \:\:t\right)}{\alpha \:\:\:\:\:\:}\right]_0^{\infty }+\frac{s}{\alpha \:}\cdot \frac{\alpha }{\alpha \:^2+s^2} L{cos(αt)}=(0−0)+sα2+s2=sα2+s2L\left\{cos\left(\alpha \:t\right)\right\}=\left(0-0\right)+\frac{s}{\alpha \:^2+s^2}=\frac{s}{\alpha \:\:^2+s^2}

Proof #7

L{α⋅f(t)}=∫0∞(αe−st)dt=α∫0∞(e−st)dt=αL{f(t)}L\left\{\alpha \cdot f\left(t\right)\right\}=\int _0^{\infty }\left(\alpha e^{-st}\right)dt=\:\alpha \int _0^{\infty }\left(\:e^{-st}\right)dt=\alpha \:L\left\{f\left(t\right)\right\}

Proof #8

L{f1(t)+f2(t)}=∫0∞((f1(t)+f2(t))e−st)dtL\left\{f_1\left(t\right)+f_2\left(t\right)\right\}=\int _0^{\infty }\left(\left(f_1\left(t\right)+f_2\left(t\right)\right)e^{-st}\right)dt L{f1(t)+f2(t)}=∫0∞(f1(t)e−st+f2(t)e−st)dtL\left\{f_1\left(t\right)+f_2\left(t\right)\right\}=\:\int _0^{\infty \:}\left(f_1\left(t\right)e^{-st}+f_2\left(t\right)e^{-st}\right)dt L{f1(t)+f2(t)}=L{f1(t)}+L{f2(t)}L\left\{f_1\left(t\right)+f_2\left(t\right)\right\}=L\left\{f_1\left(t\right)\right\}+L\left\{f_2\left(t\right)\right\}

Proof #9

L{ddtf(t)}=∫0∞ddtf(t)⋅e−stdtL\left\{\frac{d}{dt}f\left(t\right)\right\}=\int _0^{\infty }\frac{d}{dt}f\left(t\right)\cdot e^{-st}dt u=e−stu=e^{-st} dv=ddtf(t)dv=\frac{d}{dt}f\left(t\right) du=−se−stdtdu=-se^{-st}dt v=f(t)v=f\left(t\right) L{ddtf(t)}=[e−stf(t)]0∞−(∫0∞f(t)⋅(−se−st)dt)L\left\{\frac{d}{dt}f\left(t\right)\right\}=\left[e^{-st}f\left(t\right)\right]_0^{\infty }-\left(\int _0^{\infty }f\left(t\right)\cdot \left(-se^{-st}\right)dt\right) L{ddtf(t)}=[e−stf(t)]0∞−(−s∫0∞f(t)⋅(e−st)dt)L\left\{\frac{d}{dt}f\left(t\right)\right\}=\left[e^{-st}f\left(t\right)\right]_0^{\infty \:}-\left(-s\int _0^{\infty \:}f\left(t\right)\cdot \left(e^{-st}\right)dt\right) L{ddtf(t)}=(0−f(0))−(−sL{f(t)})L\left\{\frac{d}{dt}f\left(t\right)\right\}=\left(0-f\left(0\right)\right)-\left(-sL\left\{f\left(t\right)\right\}\right) L{ddtf(t)}=sL{f(t)}−f(0)L\left\{\frac{d}{dt}f\left(t\right)\right\}=sL\left\{f\left(t\right)\right\}-f\left(0\right)

Proof #10

L{d2dt2f(t)}L\left\{\frac{d^2}{dt^2}f\left(t\right)\right\} g(t)=ddtf(t)g\left(t\right)=\frac{d}{dt}f\left(t\right) ddtg(t)=d2dt2f(t)\frac{d}{dt}g\left(t\right)=\frac{d^2}{dt^2}f\left(t\right) L{d2dt2f(t)}=L{ddtg(t)}=sL{g(t)}−f(0)L\left\{\frac{d^2}{dt^2}f\left(t\right)\right\}=L\left\{\frac{d}{dt}g\left(t\right)\right\}=sL\left\{g\left(t\right)\right\}-f\left(0\right) L{d2dt2f(t)}=sL{ddtf(t)}−ddtg(0)L\left\{\frac{d^2}{dt^2}f\left(t\right)\right\}=sL\left\{\frac{d}{dt}f\left(t\right)\right\}-\frac{d}{dt}g\left(0\right) L{d2dt2f(t)}=s(sL{f(t)}−f(0))−ddtf(0)L\left\{\frac{d^2}{dt^2}f\left(t\right)\right\}=s\left(sL\left\{f\left(t\right)\right\}-f\left(0\right)\right)-\frac{d}{dt}f\left(0\right) L{d2dt2f(t)}=s2L{f(t)}−sf(0)−ddtf(0)L\left\{\frac{d^2}{dt^2}f\left(t\right)\right\}=s^2L\left\{f\left(t\right)\right\}-sf\left(0\right)-\frac{d}{dt}f\left(0\right)

Proof #11

L{∫0tf(x)dx}=∫0∞∫0tf(x)dx⋅e−stdtL\left\{\int _0^tf\left(x\right)dx\right\}=\int _0^{\infty }\int _0^tf\left(x\right)dx\cdot e^{-st}dt u=∫0tf(x)dxu=\int _0^tf\left(x\right)dx dv=e−stdtdv=e^{-st}dt du=f(x)dtdu=f\left(x\right)dt v=e−st−sv=\frac{e^{-st}}{-s} L{∫0tf(x)dx}=[∫0tf(x)dx⋅e−st−s]0∞−(∫0∞e−st−sf(x)dt)L\left\{\int _0^tf\left(x\right)dx\right\}=\left[\int _0^tf\left(x\right)dx\cdot \frac{e^{-st}}{-s}\right]_0^{\infty }-\left(\int _0^{\infty }\frac{e^{-st}}{-s}f\left(x\right)dt\:\right) L{∫0tf(x)dx}=(0−0)−1−s(∫0∞e−stf(x)dt)L\left\{\int _0^tf\left(x\right)dx\right\}=\left(0-0\right)-\frac{1}{-s}\left(\int _0^{\infty \:\:}e^{-st}f\left(x\right)dt\:\right) L{∫0tf(x)dx}=L{f(x)}sL\left\{\int _0^tf\left(x\right)dx\right\}=\frac{L\left\{f\left(x\right)\right\}}{s}