hcju/proofwikiqa huggingface dataset:
From Bessel Function of the First Kind of Order Zero: {{begin-eqn}} {{eqn | l = \map {J_0} t       | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k}       | c =  }} {{eqn | r = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \laptrans {\map {J_0} t}       | r = \laptrans {\sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k} }       | c =  }} {{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \laptrans {t^{2 k} }       | c = Linear Combination of Laplace Transforms }} {{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{2 k} \paren {k!}^2} \dfrac {\paren {2 k}!} {s^{2 k + 1} }       | c = Laplace Transform of Positive Integer Power }} {{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }       | c = {{Defof|Binomial Coefficient}}: $\dbinom {2 k} k = \dfrac {\paren {2 k}!} {\paren {k!}^2}$ }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \dfrac 1 {\sqrt {s^2 + 1} }       | r = \dfrac 1 {\sqrt {s^2} \sqrt {1 + \paren {1 / s}^2} }       | c =  }} {{eqn | r = \dfrac 1 s \paren {1 + \paren {\dfrac 1 s}^2}^{-1/2}       | c = {{Defof|Rational Power}} }} {{eqn | r = \dfrac 1 s \sum_{k \mathop = 0}^\infty \dbinom {-1/2} k \paren {\dfrac 1 s}^{2 k}       | c = General Binomial Theorem }} {{eqn | r = \dfrac 1 s \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {4^k} \dbinom {2 k} k \paren {\dfrac 1 s}^{2 k}       | c = Binomial Coefficient of Minus Half }} {{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \dfrac 1 {2^{2 k} } \dbinom {2 k} k \paren {\dfrac 1 {s^{2 k + 1} } }       | c = rearranging, and bringing $\dfrac 1 s$ inside the summation }} {{eqn | r = \sum_{k \mathop = 0}^\infty \paren {-1}^k \paren {\dfrac 1 2}^{2 k} \dbinom {2 k} k \dfrac 1 {s^{2 k + 1} }       | c = further rearrangement }} {{end-eqn}} The two expressions match, and the result follows. {{qed}}
From Bessel Function of the First Kind of Order Zero: {{begin-eqn}} {{eqn | l = \map {J_0} t       | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {\paren {k!}^2} \paren {\dfrac t 2}^{2 k}       | c =  }} {{eqn | r = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \laptrans {\map {J_0} t}       | r = \dfrac 1 s - \dfrac 1 {2^2} \dfrac {2!} {s^3} + \dfrac 1 {2^2 4^2} \dfrac {4!} {s^5} - \dfrac 1 {2^2 4^2 6^2} \dfrac {6!} {s^7} + \dotsb       | c = Laplace Transform of Positive Integer Power }} {{eqn | r = \dfrac 1 s \paren {1 - \dfrac 1 2 \paren {\dfrac 1 {s^2} } + \dfrac {1 \times 3} {2 \times 4} \paren {\dfrac 1 {s^4} } - \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \paren {\dfrac 1 {s^6} } + \dotsb}       | c = simplifying }} {{eqn | r = \dfrac 1 s \paren {\paren {1 - \dfrac 1 {s^2} } ^{-1/2} }       | c = General Binomial Theorem }} {{eqn | r = \dfrac 1 {\sqrt {s^2 + 1} }       | c = simplifying }} {{end-eqn}} {{qed}}
By definition of Bessel function of the first kind, $\map {J_0} t$ satisfies Bessel's equation: {{begin-eqn}} {{eqn | l = t^2 \, \map {\dfrac {\d^2} {\d t^2} } {\map {J_0} t} + t \, \map {\dfrac \d {\d t} } {\map {J_0} t} + \paren {t^2 - 0^2} {\map {J_0} t}       | r = 0       | c =  }} {{eqn | n = 1       | ll= \leadsto       | l = t \, {\map {J_0''} t} + \map {J_0'} t + t \, \map {J_0} t       | r = 0       | c =  }} {{end-eqn}} From Laplace Transform of Derivative: :$\laptrans {\map {J_0'} t} = s \laptrans {\map {J_0} t} - \map {J_0} 0$ and from Laplace Transform of Second Derivative: :$\laptrans {\map {J_0''} t} = s^2 \laptrans {\map {J_0} t} - s \, \map {J_0} 0 - \map {J_0'} 0$ From Bessel Function of the First Kind of Order $0$: :$\map {J_0} t = 1 - \dfrac {t^2} {2^2} + \dfrac {t^4} {2^2 \times 4^2} - \dfrac {t^6} {2^2 \times 4^2 \times 6^2} + \dotsb$ from which it follows immediately that: :$\map {J_0} 0 = 1$ From Derivative of Bessel Function of the First Kind of Order $0$ we have: {{ProofWanted|Derivative of Bessel Function of the First Kind of Order 0}} from which it follows that: :$\map {J'_0} 0 = 0$ Then from Derivative of Laplace Transform: {{begin-eqn}} {{eqn | l = \laptrans {t \, \map {J_0''} t}       | r = -\map {\dfrac \d {\d s} } {\map {J_0''} t}       | c =  }} {{eqn | r = -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s \, \map {J_0} 0 - \map {J_0'} 0}       | c =  }} {{eqn | r = -\map {\dfrac \d {\d s} } {s^2 \, \map {J_0} t - s}       | c = $\map {J_0} 0 = 1$ and $\map {J'_0} 0 = 0$ }} {{end-eqn}} and: :$\laptrans {t \map {J_0} t} = -\map {\dfrac \d {\d s} } {\map {J_0} t}$ Thus by taking the Laplace transform of $(1)$ we have: Let $y = \laptrans {\map {J_0} t}$. {{begin-eqn}} {{eqn | l = \laptrans {t \, {\map {J_0''} t} + \map {J_0'} t + t \, \map {J_0} t}       | r = \laptrans 0       | c =  }} {{eqn | ll= \leadsto       | l = -\map {\dfrac \d {\d s} } {s^2 \, \laptrans {\map {J_0} t} - s} + \paren {s \laptrans {\map {J_0} t} - \map {J_0} 0} - \map {\dfrac \d {\d s} } {\map {J_0} t}       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = -\map {\dfrac \d {\d s} } {s^2 y - s} + \paren {s y - 1} - \dfrac {\d y} {\d s}       | r = 0       | c = setting $y = \laptrans {\map {J_0} t}$ and noting that $\map {J_0} 0 = 1$ from above }} {{eqn | ll= \leadsto       | l = \dfrac {\d y} {\d s}       | r = -\dfrac {s y} {s^2 + 1}       | c = rearranging and simplifying }} {{eqn | ll= \leadsto       | l = \int \dfrac {\d y} y       | r = -\int \dfrac {s \rd s} {s^2 + 1}       | c = Separation of Variables }} {{eqn | ll= \leadsto       | l = y       | r = \dfrac c {\sqrt {s^2 + 1} }       | c = for some constant $c$ }} {{end-eqn}} Now we have that: {{begin-eqn}} {{eqn | l = \lim_{s \mathop \to \infty} s \, \map y s       | r = \lim_{s \mathop \to \infty} \dfrac {c s} {\sqrt {s^2 + 1} }       | c =  }} {{eqn | r = c       | c =  }} {{end-eqn}} and: :$\displaystyle \lim_{t \mathop \to 0} \map {J_0} t = 1$ Thus by the Initial Value Theorem of Laplace Transform: :$c = 1$ and so: :$\laptrans {t \, \map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$ {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {t^2 \frac {\d^2 x} {\d t^2} + t \frac {\d x} {\d t} + (t^2-\alpha^2)x} s        | r = 0        | c = Laplace Transform of Bessel's Equation }} {{eqn | l = \frac {\d^2} {\d s^2}\mathcal{L}[x''] - \frac \d {\d s} \mathcal{L}[x'] + \frac {\d^2} {\d s^2} \mathcal{L}[x] - \alpha^2 \mathcal{L} [x]       | r = 0       | c = setting the initial conditions as $x(0)=1,\ x'(0)=0$ }} {{eqn | l = (s^2+1) \mathcal{L}''[x] + 3 s \mathcal {L}'[x] + (1 - \alpha^2) \mathcal{L}[x]       | r = 0       | c = Make the following change of variable $u = \sqrt {s^2 + 1} \mathcal {L} [x]$ }} {{eqn | l = u'' \sqrt {s^2 + 1} + \dfrac s {\sqrt {s^2 + 1} } u'       | r = \frac {\alpha^2 u} {\sqrt {s^2 + 1} }       | c =  }} {{eqn | l = (u' \sqrt {s^2 + 1} )'       | r = \frac {\alpha^2 u} {\sqrt {s^2 + 1} }       | c = multiplying both sides by $u' \sqrt {s^2 + 1}$ }} {{eqn | l = \frac 1 2 ( (u' \sqrt {s^2 + 1} )^2)'       | r = \frac 1 2 \alpha^2 (u^2)'       | c =  }} {{eqn | l = u' \sqrt {s^2 + 1}       | r = -\alpha u       | c = Constant of Integration removed by the Final Value Theorem of Laplace Transform }} {{eqn | l = \int \frac 1 u \rd u       | r = \int -\frac \alpha {\sqrt {s^2 + 1} } \rd s       | c =  }} {{eqn | l = \map \ln u       | r = \alpha \ln (\sqrt {s^2 + 1} - s)       | c =  }} {{eqn | l = u       | r = (\sqrt {s^2 + 1} - s)^\alpha       | c = reverting the substitution }} {{eqn | l = \mathcal{L} [x]       | r = \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }       | c =  }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \mathcal{L}[J_\alpha (t)](s)       | r = \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }       | c =  }} {{end-eqn}} {{qed}}
By Derivative of Error Function, we have:  :$\displaystyle \frac \d {\d t} \paren {\map \erf t} = \frac 2 {\sqrt \pi} e^{-t^2}$ By Primitive of Exponential Function, we have:  :$\displaystyle \int e^{-s t} \rd t = -\frac {e^{-s t} } s$ So:  {{begin-eqn}} {{eqn	| l = \laptrans {\map \erf t} 	| r = \int_0^\infty e^{-s t} \map \erf t \rd t 	| c = {{Defof|Laplace Transform}} }} {{eqn	| r = \intlimits {-\frac 1 s e^{-s t} \map \erf t} 0 \infty - \int_0^\infty \paren {-\frac 2 {\sqrt \pi} \frac {e^{-s t} } s e^{-t^2} } \rd t 	| c = Integration by Parts }} {{eqn	| r = -\frac 1 s \lim_{t \mathop \to \infty} \paren {e^{-s t} \map \erf t} + \frac 1 s e^0 \erf 0 + \frac 2 {s \sqrt \pi} \int_0^\infty \exp \paren {-s t - t^2} \rd t }} {{end-eqn}} We have:  {{begin-eqn}} {{eqn	| l = \lim_{t \mathop \to \infty} \paren {e^{-s t} \map \erf t} 	| r = \paren {\lim_{t \mathop \to \infty} e^{-s t} } \paren {\lim_{t \mathop \to \infty} \map \erf t} 	| c = Combination Theorem for Limits of Functions: Product Rule }} {{eqn	| r = 0 \times 1 	| c = Exponential Tends to Zero and Infinity, Limit to Infinity of Error Function }} {{eqn	| r = 0 }} {{end-eqn}} We also have:  {{begin-eqn}} {{eqn	| l = \frac 1 s e^0 \erf 0 	| r = \frac 1 s \int_0^0 e^{-t^2} \rd t 	| c = Exponential of Zero, {{Defof|Error Function}} }} {{eqn	| r = 0 	| c = Definite Integral on Zero Interval }} {{end-eqn}} Therefore:  {{begin-eqn}} {{eqn	| l = \laptrans {\map \erf t}  	| r = \frac 2 {s \sqrt \pi} \int_0^\infty \map \exp {-\paren {t^2 + s t} } \rd t }} {{eqn	| r = \frac 2 {s \sqrt \pi} \int_0^\infty \map \exp {-\paren {\paren {t + \frac s 2}^2 - \frac {s^2} 4} } \rd t 	| c = completing the square }} {{eqn	| r = \frac 2 {s \sqrt \pi} \map \exp {\frac {s^2} 4} \int_0^\infty \map \exp {-\paren {t + \frac s 2}^2} \rd t 	| c = Exponential of Sum }} {{eqn	| r = \frac 2 {s \sqrt \pi} \map \exp {\frac {s^2} 4} \int_{\frac s 2}^\infty \map \exp {-u^2} \rd u 	| c = substituting $u = t + \dfrac s 2$ }} {{eqn	| r = \frac 1 s \map \exp {\frac {s^2} 4} \paren {\frac 2 {\sqrt \pi} \int_{\frac s 2}^\infty \map \exp {-u^2} \rd u} }} {{eqn	| r = \frac 1 s \map \exp {\frac {s^2} 4} \map \erfc {\frac s 2} 	| c = {{Defof|Complementary Error Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\map \erf {\sqrt t} }       | r = \laptrans {\frac 2 {\sqrt \pi} \int_0^{\sqrt t} \map \exp {-u^2} \rd u}       | c = {{Defof|Error Function}} }} {{eqn | r = \laptrans {\frac 2 {\sqrt \pi} \int_0^{\sqrt t} \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {u^{2 n} } {n!} \rd u} }       | c = {{Defof|Exponential Function/Real|Real Exponential Function|subdef = Sum of Series}} }} {{eqn | r = \laptrans {\frac 2 {\sqrt \pi} \paren {\sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {\sqrt t}^{2 n + 1} } {\paren {2 n + 1} n!} } }       | c = Primitive of Power }} {{eqn | r = \frac 2 {\sqrt \pi} \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1} n!} \laptrans {t^{n + \frac 1 2} }       | c = Linear Combination of Laplace Transforms }} {{eqn | r = \frac 2 {\sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 3 2} } {\paren {2 n + 1} n! s^{n + \frac 3 2} }        | c = Laplace Transform of Real Power }} {{eqn | r = \frac 2 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {n + \frac 1 2} \map \Gamma {n + \frac 1 2} } {\paren {2 n + 1} n! s^n}       | c = Gamma Difference Equation }} {{eqn | r = \frac 1 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 1 2} } {\map \Gamma {n + 1} s^n}        | c = Gamma Function Extends Factorial }} {{end-eqn}} We have:  {{begin-eqn}} {{eqn | l = \map \Gamma {n + \frac 1 2}       | r = \frac \pi {\map \sin {\pi \paren {\frac 1 2 - n} } \map \Gamma {\frac 1 2 - n} }       | c = Euler's Reflection Formula }} {{eqn | r = \frac \pi {\map \cos {-n \pi} \map \Gamma {\frac 1 2 - n} }       | c = Sine of Complement equals Cosine }} {{eqn | r = \frac {\paren {-1}^n \pi} {\map \Gamma {\frac 1 2 - n} }       | c = Cosine of Integer Multiple of Pi }} {{end-eqn}} So:  {{begin-eqn}} {{eqn | l = \frac 1 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 1 2} } {\map \Gamma {n + 1} s^n}        | r = \frac 1 {s^{3/2} \sqrt \pi} \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {-1}^n \pi} {\map \Gamma {n + 1} \map \Gamma {\frac 1 2 - n} s^n}       | c =  }} {{eqn | r = \frac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \frac {\sqrt \pi} {\map \Gamma {n + 1} \map \Gamma {\frac 1 2 - n} s^n}       | c =  }} {{eqn | r = \frac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \frac {\map \Gamma {-\frac 1 2 + 1} } {\map \Gamma {n + 1} \map \Gamma {-\frac 1 2 - n + 1} s^n}       | c = Gamma Function of One Half }} {{eqn | r = \dfrac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \binom {-\frac 1 2} n \frac 1 {s^n}       | c = {{Defof|Binomial Coefficient/Complex Numbers|Binomial Coefficient}} }} {{eqn | r = \dfrac 1 {s^{3/2} } \paren {1 + \dfrac 1 s}^{-1/2}       | c = General Binomial Theorem }} {{eqn | r = \dfrac 1 {s \sqrt {s + 1} }       | c = simplification }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\map {u_c} t}       | r = \int_0^{\to +\infty} \map {u_c} t e^{-s t} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \int_0^c \map {u_c} t e^{-s t} \rd t + \int_c^{\to +\infty} \map {u_c} t e^{-s t} \rd t       | c = Sum of Integrals on Adjacent Intervals for Integrable Functions }} {{eqn | r = \int_0^c 0 \times e^{-s t} \rd t + \int_c^{\to +\infty} 1 \times e^{-s t} \rd t       | c = {{Defof|Heaviside Step Function}} }} {{eqn | r = \int_c^{\to +\infty} e^{-s t} \rd t       | c =  }} {{eqn | r = \lim_{L \mathop \to +\infty} \int_c^L e^{-s t} \rd t       | c = {{Defof|Improper Integral}} }} {{eqn | r = \lim_{L \mathop \to +\infty} \intlimits {\dfrac {e^{-s t} } {-s} } c L       | c = Primitive of $e^{a x}$ }} {{eqn | r = \lim_{L \mathop \to +\infty} \dfrac {e^{-s L} } {-s} - \dfrac {e^{-s c} } {-s}       | c =  }} {{eqn | r = 0 + \dfrac {e^{-s c} } s       | c = simplification }} {{end-eqn}} Hence the result. {{qed}}
=== Lemma === {{:Laplace Transform of Dirac Delta Function/Lemma}} Then: {{begin-eqn}} {{eqn | l = \laptrans {\map \delta t}       | r = \lim_{\epsilon \mathop \to 0} \laptrans {\map {F_\epsilon} t}       | c = {{Defof|Dirac Delta Function}} }} {{eqn | r = \lim_{\epsilon \mathop \to 0} \dfrac {1 - e^{-s \epsilon} } {\epsilon s}       | c = Lemma }} {{eqn | r = \lim_{\epsilon \mathop \to 0} \dfrac 1 {\epsilon s} \paren {1 - \paren {1 - s \epsilon + \dfrac {s^2 \epsilon^2} {2!} - \dotsb} }       | c = {{Defof|Exponential Function/Real|subdef = Sum of Series|Exponential Function}} }} {{eqn | r = \lim_{\epsilon \mathop \to 0} \paren {1 - \dfrac {s \epsilon} {2!} + \dotsb}       | c =  }} {{eqn | r = 1       | c =  }} {{end-eqn}} {{qed}}
=== Lemma === {{:Laplace Transform of Dirac Delta Function/Lemma}} Then: {{begin-eqn}} {{eqn | l = \laptrans {\map \delta t}       | r = \lim_{\epsilon \mathop \to 0} \laptrans {\map {F_\epsilon} t}       | c = {{Defof|Dirac Delta Function}} }} {{eqn | r = \lim_{\epsilon \mathop \to 0} \dfrac {1 - e^{-s \epsilon} } {\epsilon s}       | c = Lemma }} {{eqn | r = \lim_{\epsilon \mathop \to 0} \dfrac {\map {\dfrac \d {\d s} } {1 - e^{-s \epsilon} } } {\map {\dfrac \d {\d s} } {\epsilon s} }       | c = L'Hôpital's Rule }} {{eqn | r = \lim_{\epsilon \mathop \to 0} \dfrac {\paren {-\epsilon} \times \paren {-e^{-s \epsilon} } } \epsilon       | c =  }} {{eqn | r = \lim_{\epsilon \mathop \to 0} e^{-s \epsilon}       | c = simplification }} {{eqn | r = 1       | c = Exponential of Zero }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\map \delta {t - a} }       | r = \int_0^{\to +\infty} e^{-s t} \map \delta {t - a} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = e^{-s \times a}       | c = Integral to Infinity of Shifted Dirac Delta Function by Continuous Function }} {{eqn | r = e^{-a s}       | c =  }} {{end-eqn}} {{qed}}
From Limit of Sine of X over X: :$\displaystyle \lim_{x \mathop \to 0} \frac {\sin x} x = 1$ From Laplace Transform of Sine: :$(1): \quad \displaystyle \laptrans {\sin t} = \dfrac 1 {s^2 + 1}$ From Laplace Transform of Integral: :$(2): \quad \displaystyle \laptrans {\dfrac {\map f t} t} = \int_s^{\to \infty} \map F u \rd u$ Hence: {{begin-eqn}} {{eqn | l = \laptrans {\dfrac {\sin t} t}       | r = \int_s^{\to \infty} \dfrac 1 {u^2 + 1} \rd u       | c = $(1)$ and $(2)$ }} {{eqn | r = \lim_{L \mathop \to \infty} \int_s^L \dfrac 1 {u^2 + 1} \rd u       | c = {{Defof|Improper Integral}} }} {{eqn | r = \lim_{L \mathop \to \infty} \bigintlimits {\arctan u} s L       | c = Primitive of $\dfrac 1 {x^2 + a^2}$ }} {{eqn | r = \lim_{L \mathop \to \infty} \paren {\arctan L - \arctan s}       | c =  }} {{eqn | r = \dfrac \pi 2 - \arctan s       | c =  }} {{eqn | r = \arccot s       | c = Sum of Arctangent and Arccotangent }} {{eqn | r = \arctan \dfrac 1 s       | c = Arctangent of Reciprocal equals Arccotangent }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\dfrac {\sin t} t}       | r = \arctan \dfrac 1 s       | c = Laplace Transform of Sine of t over t }} {{eqn | ll= \leadsto       | l = \laptrans {\dfrac {\sin a t} {a t} }       | r = \dfrac 1 a \arctan \dfrac 1 {s / a}       | c = Laplace Transform of Constant Multiple }} {{eqn | r = \dfrac 1 a \arctan \dfrac a s       | c =  }} {{end-eqn}} But: {{begin-eqn}} {{eqn | l = \laptrans {\dfrac {\sin a t} {a t} }       | r = \int_0^\infty e^{-s t} \dfrac {\sin a t} {a t} \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \dfrac 1 a \int_0^\infty e^{-s t} \dfrac {\sin a t} t \rd t       | c =  }} {{eqn | r = \dfrac 1 a \laptrans {\dfrac {\sin a t} t}       | c =  }} {{end-eqn}} The result follows. {{Qed}}
{{begin-eqn}} {{eqn | l = \laptrans {t \sin a t}       | r = -\map {\dfrac \d {\d s} } {\laptrans {\sin a t} }       | c = Derivative of Laplace Transform }} {{eqn | r = -\map {\dfrac \d {\d s} } {\dfrac a {s^2 + a^2} }       | c = Laplace Transform of Sine }} {{eqn | r = \dfrac {2 a s} {\paren {s^2 + a^2}^2}       | c = Quotient Rule for Derivatives }} {{end-eqn}} {{qed}}
We have: {{begin-eqn}} {{eqn | l = \laptrans {\cos a t}       | r = \int_0^\infty e^{-s t} \cos a t \rd t       | c = {{Defof|Laplace Transform}} }} {{eqn | n = 1       | r = \dfrac s {s^2 + a^2}       | c = Laplace Transform of Cosine }} {{end-eqn}} Hence: {{begin-eqn}} {{eqn | l = \dfrac \d {\d a} \int_0^\infty e^{-s t} \cos a t \rd t       | r = \int_0^\infty e^{-s t} \paren {\map {\dfrac \partial {\partial a} } {\cos a t} } \rd t       | c = Derivative of Integral }} {{eqn | r = \int_0^\infty e^{-s t} \paren {-t \sin a t} \rd t       | c = Derivative of Cosine Function }} {{eqn | r = -\laptrans {t \sin a t}       | c = {{Defof|Laplace Transform}} }} {{eqn | r = \map {\dfrac \d {\d a} } {\dfrac s {s^2 + a^2} }       | c = from $(1)$ }} {{eqn | r = -\dfrac {2 a s} {\paren {s^2 + a^2}^2}       | c = Quotient Rule for Derivatives }} {{end-eqn}} and the result follows. {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {t^2 \cos a t}       | r = -\map {\dfrac {\d^2} {\d s^2} } {\laptrans {\cos a t} }       | c = Higher Order Derivatives of Laplace Transform }} {{eqn | r = -\map {\dfrac {\d^2} {\d s^2} } {\dfrac a {s^2 + a^2} }       | c = Laplace Transform of Sine }} {{eqn | r = \dfrac {2 s^3 - 6 a^2 s} {\paren {s^2 + a^2}^3}       | c = Quotient Rule for Derivatives }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \laptrans {\dfrac {\map f t} t}       | r = \int_s^{\to \infty} \map F u \rd u       | c = Integral of Laplace Transform }} {{eqn | ll= \leadsto       | l = \int_0^\infty e^{-s t} {\dfrac {\map f t} t} \rd t       | r = \int_s^{\to \infty} \map F u \rd u       | c = {{Defof|Laplace Transform}} }} {{eqn | ll= \leadsto       | l = \lim_{s \mathop \to 0} \int_0^\infty e^{-s t} \dfrac {\map f t} t \rd t       | r = \lim_{s \mathop \to 0} \int_s^{\to \infty} \map F u \rd u       | c =  }} {{eqn | ll= \leadsto       | l = \int_0^\infty \dfrac {\map f t} t \rd t       | r = \int_0^{\to \infty} \map F u \rd u       | c =  }} {{end-eqn}} {{qed}}
We have that $\map f t$ is periodic with period $2 \pi$: :800px Hence: {{begin-eqn}} {{eqn | l = \laptrans {\map f t}       | r = \dfrac 1 {1 - e^{-2 \pi s} } \int_0^{2 \pi} e^{-s t} \map f t \rd t       | c = Laplace Transform of Periodic Function }} {{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\int_0^\pi e^{-s t} \sin t \rd t + \int_\pi^{2 \pi} e^{-s t} \times 0 \rd t}       | c =  }} {{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \int_0^\pi e^{-s t} \sin t \rd t       | c =  }} {{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \intlimits {\dfrac {e^{-s t} \paren {-s \sin t - \cos t} } {s^2 + 1} } 0 \pi       | c = Primitive of $e^{a x} \sin b x$ }} {{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-s \sin \pi - \cos \pi} } {s^2 + 1} - \dfrac {e^{-s \times 0} \paren {-s \sin 0 - \cos 0} } {s^2 + 1} }       | c =  }} {{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-\cos \pi} } {s^2 + 1} - \dfrac {e^{-s \times 0} \paren {- \cos 0} } {s^2 + 1} }       | c = Sine of Integer Multiple of Pi }} {{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-\paren {-1} } } {s^2 + 1} - \dfrac {e^{-s \times 0} \paren {-1} } {s^2 + 1} }       | c = Cosine of Integer Multiple of Pi }} {{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} \paren {-\paren {-1} } } {s^2 + 1} - \dfrac {1 \times \paren {-1} } {s^2 + 1} }       | c = Exponential of Zero }} {{eqn | r = \dfrac 1 {1 - e^{-2 \pi s} } \paren {\dfrac {e^{-s \pi} + 1} {s^2 + 1} }       | c = simplification }} {{eqn | r = \dfrac 1 {\paren {1 + e^{-\pi s} } \paren {1 - e^{-\pi s} } } \paren {\dfrac {e^{-s \pi} + 1} {s^2 + 1} }       | c = Difference of Two Squares }} {{eqn | r = \dfrac 1 {\paren {1 - e^{-\pi s} } \paren {s^2 + 1} }       | c = simplification }} {{end-eqn}} {{qed}}
Let $\displaystyle I = \int_0^\infty \map \exp {-t^2} \rd t$. Let $\displaystyle I_P = \int_0^P \map \exp {-x^2} \rd x = \int_0^P \map \exp {-y^2} \rd y$. Then we have: :$I = \displaystyle \lim_{P \mathop \to \infty} I_P$ Hence: {{begin-eqn}} {{eqn | l = {I_P}^2       | r = \paren {\int_0^P \map \exp {-x^2} \rd x} \paren {\int_0^P \map \exp {-y^2} \rd y}       | c =  }} {{eqn | r = \int_0^P \int_0^P \map \exp {-\paren {x^2 + y^2} } \rd x \rd y       | c =  }} {{eqn | r = \iint_{\mathscr R_P} \map \exp {-\paren {x^2 + y^2} } \rd x \rd y       | c =  }} {{end-eqn}} where $\mathscr R_P$ is the square $\Box OACE$ in the figure below: :400px Because the integrand is positive: {{begin-eqn}} {{eqn | n = 1       | l = \iint_{\mathscr R_1} \map \exp {-\paren {x^2 + y^2} } \rd x \rd y       | o = \le       | r = {I_P}^2       | c =  }} {{eqn | o = \le       | r = \iint_{\mathscr R_2} \map \exp {-\paren {x^2 + y^2} } \rd x \rd y       | c =  }} {{end-eqn}} where $\mathscr R_1$ and $\mathscr R_2$ are the regions in the first quadrant bounded by the circles with centers $O$ and radii $P$ and $P \sqrt 2$ respectively. Using polar coordinates, we can express $(1)$ as: {{begin-eqn}} {{eqn | n = 1       | l = \int_{\theta \mathop = 0}^{\frac \pi 2} \int_{r \mathop = 0}^P \map \exp {-r^2} r \rd r \rd \theta       | o = \le       | r = {I_P}^2       | c =  }} {{eqn | o = \le       | r = \int_{\theta \mathop = 0}^{\frac \pi 2} \int_{r \mathop = 0}^{P \sqrt 2} \map \exp {-r^2} r \rd r \rd \theta       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac \pi 4 \paren {1 - \map \exp {-P^2} }       | o = \le       | r = {I_P}^2       | c =  }} {{eqn| o = \le       | r = \dfrac \pi 4 \paren {1 - \map \exp {-2 P^2} }       | c =  }} {{eqn | ll= \leadsto       | l = \lim_{P \mathop \to \infty} {I_P}^2       | r = I^2       | c =  }} {{eqn | r = \dfrac \pi 4       | c =  }} {{eqn | ll= \leadsto       | l = I       | r = \dfrac {\sqrt \pi} 2       | c =  }} {{end-eqn}} {{qed}}