Proof #1

L\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\left\{t\right\}=\int _0^{\infty \:}\left(t\cdot \:e^{-st}\right)dt u=t dv=e^{-st} du=dt v=\frac{1}{-s}e^{-st} L\left\{t\right\}=\left[t\cdot \:\:\:\frac{1}{-s}e^{-st}\right]^{\infty \:}_0-\int \:\:\left(\frac{1}{-s}e^{-st}\right)dt L\left\{t\right\}=\left(0-0\right)-\left(\frac{1}{-s}\right)\int \:\:\:\left(e^{-st}\right)dt 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\left\{t\right\}=\frac{1}{s}\cdot \frac{1}{s}=\frac{1}{s^2}

Proof #3

L\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\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\left\{e^{-\alpha \:t}\right\}=\frac{1}{\alpha +s}

Proof #4

L\left\{sin\left(\alpha \:t\right)\right\}=\int _0^{\infty \:}\left(sin\left(\alpha \:\:t\right)\cdot \:\:e^{-st}\right)dt u=e^{-st} dv=\:sin\left(\alpha t\right)dt du=-se^{-st}dt v=\frac{-cos\left(\alpha \:t\right)}{\alpha \:} \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) \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^{-st} dv=\:cos\left(\alpha t\right)dt du=-se^{-st}dt v=\frac{sin\left(\alpha t\right)}{\alpha } \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\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 \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}   \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} \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\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\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\left\{cos\left(\alpha t\right)\right\}=\int _0^{\infty }\left(cos\left(\alpha t\right)\cdot e^{-st}\right)dt u=e^{-st} dv=\:cos\left(\alpha t\right)dt du=-se^{-st}dt v=\frac{sin\left(\alpha t\right)}{\alpha } L\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\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\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\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\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\left\{te^{-\alpha t}\right\}=\int _0^{\infty }te^{-\alpha t}e^{-st}=\int _0^{\infty }te^{\left(-\alpha -s\right)t} u=t dv=e^{\left(-\alpha -s\right)t}dt du=1dt v=\frac{1}{\left(-\alpha \:\:-s\right)}e^{\left(-\alpha \:-s\right)t} L\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\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\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\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\left\{cos\left(\alpha t\right)\right\}=\int _0^{\infty }\left(cos\left(\alpha t\right)\cdot e^{-st}\right)dt u=e^{-st} dv=\:cos\left(\alpha t\right)dt du=-se^{-st}dt v=\frac{sin\left(\alpha t\right)}{\alpha } L\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\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\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\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\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\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\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\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\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\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^{-st} dv=\frac{d}{dt}f\left(t\right) du=-se^{-st}dt v=f\left(t\right) 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\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\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\left\{\frac{d}{dt}f\left(t\right)\right\}=sL\left\{f\left(t\right)\right\}-f\left(0\right)

Proof #10

L\left\{\frac{d^2}{dt^2}f\left(t\right)\right\} g\left(t\right)=\frac{d}{dt}f\left(t\right) \frac{d}{dt}g\left(t\right)=\frac{d^2}{dt^2}f\left(t\right) 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\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\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\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\left\{\int _0^tf\left(x\right)dx\right\}=\int _0^{\infty }\int _0^tf\left(x\right)dx\cdot e^{-st}dt u=\int _0^tf\left(x\right)dx dv=e^{-st}dt du=f\left(x\right)dt v=\frac{e^{-st}}{-s} 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\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\left\{\int _0^tf\left(x\right)dx\right\}=\frac{L\left\{f\left(x\right)\right\}}{s}