avewright/proofwiki-math huggingface dataset:
Theorem: Derivative of x^n by Bessel Function of the First Kind of Order n of x

Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.
Then:
:$\map {\dfrac \d {\d x} } {x^n \map {J_n} x} = x^n \map {J_{n - 1} } x$

Proof:
{{begin-eqn}}
{{eqn | l = \map {\frac \d {\d x} } {x^n \map {J_n} x}
      | r = n x^{n - 1} \map {J_n} x + x^n \map {J_n'} x
      | c = Product Rule for Derivatives
}}
{{eqn | r = n x^{n - 1} \map {J_n} x - x^n \paren {\frac {\map {J_{n + 1} } x - \map {J_{n - 1} } x} 2}
      | c = Recurrence Formula for Bessel Function of the First Kind
}}
{{eqn | r = n x^{n - 1} \map {J_n} x - x^n \paren {\frac {\frac {2 n} x \map {J_n} x - 2 \map {J_{n - 1} } x} 2}
      | c = Recurrence Formula for Bessel Function of the First Kind
}}
{{eqn | r = n x^{n - 1} \map {J_n} x - x^n \paren {\frac n x \map {J_n} x - \map {J_{n - 1} } x}
}}
{{eqn | r = n x^{n - 1} \map {J_n} x - n x^{n - 1} \map {J_n} x + x^n \map {J_{n - 1} } x
}}
{{eqn | r = x^n \map {J_{n - 1} } x
}}
{{end-eqn}}
{{qed}}
Theorem: Series Expansion of Bessel Function of the First Kind/Negative Index

{{begin-eqn}}
{{eqn | l = \map {J_{-n} } x
      | r = \dfrac {x^{-n} } {2^{-n} \, \map \Gamma {1 - n} } \paren {1 - \dfrac {x^2} {2 \paren {2 - 2 n} } + \dfrac {x^4} {2 \times 4 \paren {2 - 2 n} \paren {4 - 2 n} } - \cdots}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {k + 1 - n} } \paren {\dfrac x 2}^{2 k - n}
      | c = 
}}
{{end-eqn}}

Proof:
From Series Expansion of Bessel Function of the First Kind:
{{begin-eqn}}
{{eqn | n = 1
      | l = \map {J_n} x
      | r = \dfrac {x^n} {2^n \, \map \Gamma {n + 1} } \paren {1 - \dfrac {x^2} {2 \paren {2 n + 2} } + \dfrac {x^4} {2 \times 4 \paren {2 n + 2} \paren {2 n + 4} } - \cdots}
      | c = 
}}
{{eqn | r = \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{n + 2 k}
      | c = 
}}
{{end-eqn}}
The result follows by substituting $-n$ for $n$ in $1$ and simplifying.
{{qed}}
Theorem: Integral to Infinity of Dirac Delta Function

Let $\map \delta x$ denote the Dirac delta function.
Then:
:$\ds \int_0^{+ \infty} \map \delta x \rd x = 1$

Proof:
We have that:
:$\map \delta x = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin {cases} 0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le x \le \epsilon \\ 0 & : x > \epsilon \end {cases}$
We have that:
{{begin-eqn}}
{{eqn | l = \int_0^{+ \infty} \map {F_\epsilon} x \rd x
      | r = \int_0^\epsilon \dfrac 1 \epsilon \rd x + \int_\epsilon^\infty 0 \rd x
      | c = Definition 1 of $\map {F_\epsilon} x$
}}
{{eqn | r = \int_0^\epsilon \dfrac 1 \epsilon \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \rd x
      | c = {{Defof|Improper Integral}}
}}
{{eqn | r = \dfrac 1 \epsilon \paren {\epsilon - 0} + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon} }
      | c = Definite Integral of Constant
}}
{{eqn | r = 1 + \lim_{L \mathop \to \infty} 0
      | c = simplification
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \int_0^{+ \infty} \map \delta x \rd x
      | r = \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \rd x
      | c = Definition 1 of Dirac Delta Function
}}
{{eqn | r = \lim_{\epsilon \mathop \to 0} 1
      | c = from above
}}
{{eqn | r = 1
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Integral to Infinity of Dirac Delta Function by Continuous Function

Let $\map \delta x$ denote the Dirac delta function.
Let $g$ be a continuous real function.
Then:
:$\ds \int_0^{+ \infty} \map \delta x \map g x \rd x = \map g 0$

Proof:
We have that:
:$\map \delta x = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin{cases} 0 & : x < 0 \\ \dfrac 1 \epsilon & : 0 \le x \le \epsilon \\ 0 & : x > \epsilon \end{cases}$
We have that:
{{begin-eqn}}
{{eqn | l = \int_0^{+ \infty} \map {F_\epsilon} x \map g x \rd x
      | r = \int_0^\epsilon \dfrac 1 \epsilon \map g x \rd x + \int_\epsilon^\infty 0 \times \map g x \rd x
      | c = {{Defof|Dirac Delta Function|index = 1}}
}}
{{eqn | r = \int_0^\epsilon \dfrac 1 \epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \times \map g x \rd x
      | c = {{Defof|Improper Integral}}
}}
{{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \int_\epsilon^L 0 \rd x
      | c = 
}}
{{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \epsilon} }
      | c = Definite Integral of Constant
}}
{{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x + \lim_{L \mathop \to \infty} 0
      | c = simplification
}}
{{eqn | r = \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x
      | c = 
}}
{{end-eqn}}
From Darboux's Theorem:
:$\ds m \paren {\epsilon - 0} \le \int_0^\epsilon \map g x \rd x \le M \paren {\epsilon - 0}$
where:
:$M$ is the maximum of $\map g x$
:$m$ is the minimum of $\map g x$
on $\closedint 0 \epsilon$.
Hence:
:$\ds m \epsilon \le \int_0^\epsilon \map g x \rd x \le M \epsilon$
and so dividing by $\epsilon$:
:$\ds m \le \dfrac 1 \epsilon \int_0^\epsilon \map g x \rd x \le M$
Then:
:$\ds \lim_{\epsilon \mathop \to 0} M = m = \map g 0$
and so by the Squeeze Theorem:
:$\ds \lim_{\epsilon \mathop \to 0} \int_0^{+\infty} \map {F_\epsilon} x \map g x \rd x = \map g 0$
But by {{Defof|Dirac Delta Function|index = 1}}:
:$\ds \lim_{\epsilon \mathop \to 0} \int_0^{+\infty} \map {F_\epsilon} x \map g x \rd x = \int_0^{+\infty} \map \delta x \map g x \rd x$
Hence the result.
{{qed}}
Theorem: Integral to Infinity of Shifted Dirac Delta Function by Continuous Function

Let $\map \delta x$ denote the Dirac delta function.
Let $g$ be a continuous real function.
Let $a \in \R_{\ge 0}$ be a positive real number.
Then:
:$\ds \int_0^{+ \infty} \map \delta {x - a} \, \map g x \rd x = \map g a$

Proof:
We have that:
:$\map \delta {x - a} = \ds \lim_{\epsilon \mathop \to 0} \map {F_\epsilon} x$
where:
:$\map {F_\epsilon} x = \begin {cases} 0 & : x < a \\ \dfrac 1 \epsilon & : a \le x \le a + \epsilon \\ 0 & : x > a + \epsilon \end {cases}$
We have that:
{{begin-eqn}}
{{eqn | l = \int_0^{+ \infty} \map {F_\epsilon} x \map g x \rd x
      | r = \int_0^a 0 \times \map g x + \int_a^{a + \epsilon} \dfrac 1 \epsilon \map g x \rd x + \int_{a + \epsilon}^\infty 0 \times \map g x \rd x
      | c = Definition of $\map {F_\epsilon} x$
}}
{{eqn | r = \int_0^a 0 \times \map g x + \int_a^{a + \epsilon} \dfrac 1 \epsilon \map g x \rd x + \lim_{L \mathop \to \infty} \int_{a + \epsilon}^L 0 \times \map g x \rd x
      | c = {{Defof|Improper Integral}}
}}
{{eqn | r = \int_0^a 0 + \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x + \lim_{L \mathop \to \infty} \int_{a + \epsilon}^L 0 \rd x
      | c = 
}}
{{eqn | r = \paren {0 \times \paren {a - 0} } + \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x + \lim_{L \mathop \to \infty} \paren {0 \times \paren {L - \paren {a + \epsilon} } }
      | c = Definite Integral of Constant
}}
{{eqn | r = 0 + \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x + \lim_{L \mathop \to \infty} 0
      | c = simplification
}}
{{eqn | r = \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x
      | c = 
}}
{{end-eqn}}
From Darboux's Theorem:
:$\ds m \paren {\paren {a + \epsilon} - a} \le \int_a^{a + \epsilon} \map g x \rd x \le M \paren {\paren {a + \epsilon} - 0}$
where:
:$M$ is the maximum of $\map g x$
:$m$ is the minimum of $\map g x$
on $\closedint a {a + \epsilon}$.
Hence:
:$\ds m \epsilon \le \int_a^{a + \epsilon} \map g x \rd x \le M \epsilon$
and so dividing by $\epsilon$:
:$\ds m \le \dfrac 1 \epsilon \int_a^{a + \epsilon} \map g x \rd x \le M$
Then:
:$\ds \lim_{\epsilon \mathop \to 0} M = m = \map g a$
and so by the Squeeze Theorem:
:$\ds \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \map g x \rd x = \map g a$
But by definition of the Dirac delta function:
:$\ds \lim_{\epsilon \mathop \to 0} \int_0^{+ \infty} \map {F_\epsilon} x \map g x \rd x = \int_0^{+ \infty} \map \delta {x - a} \map g x \rd x$
Hence the result.
{{qed}}
Theorem: Function which is Zero except on Countable Set of Points is Null

Let $S \subseteq \R$ be a subset of $\R$ such that $S$ is countable, either finite or countably infinite.
Let $f: \R \to \R$ be a real function such that:
:$\forall x \in \R \setminus S: \map f x = 0$
That is, except perhaps for the elements of $S$, the value of $f$ is zero.
Then $f$ is a null function.

Proof:
This is an instance of Measurable Function Zero A.E. iff Absolute Value has Zero Integral.
{{qed}}
Theorem: Laplace Transform of Bessel Function of the First Kind of Order Zero

Let $J_0$ denote the Bessel function of the first kind of order $0$.
Then the Laplace transform of $J_0$ is given as:
:$\laptrans {\map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$

Proof:
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}}
Theorem: Laplace Transform of Bessel Function of the First Kind of Order Zero

Let $J_0$ denote the Bessel function of the first kind of order $0$.
Then the Laplace transform of $J_0$ is given as:
:$\laptrans {\map {J_0} t} = \dfrac 1 {\sqrt {s^2 + 1} }$

Proof:
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}}
Theorem: Laplace Transform of Bessel Function of the First Kind

Let $J_n$ denote the Bessel function of the first kind of order $n$.
Then the Laplace transform of $J_n$ is given as:
:$\laptrans {\map {J_n} {a t} } = \dfrac {\paren {\sqrt {s^2 + a^2} - s}^n} {a^n \sqrt {s^2 + a^2} }$

Proof:
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}}
Theorem: Laplace Transform of Bessel Function of the First Kind

Let $J_n$ denote the Bessel function of the first kind of order $n$.
Then the Laplace transform of $J_n$ is given as:
:$\laptrans {\map {J_n} {a t} } = \dfrac {\paren {\sqrt {s^2 + a^2} - s}^n} {a^n \sqrt {s^2 + a^2} }$

Proof:
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}}
Theorem: Laplace Transform of Bessel Function of the First Kind

Let $J_n$ denote the Bessel function of the first kind of order $n$.
Then the Laplace transform of $J_n$ is given as:
:$\laptrans {\map {J_n} {a t} } = \dfrac {\paren {\sqrt {s^2 + a^2} - s}^n} {a^n \sqrt {s^2 + a^2} }$

Proof:
{{begin-eqn}}
{{eqn | l = \laptrans {t^2 \frac {\d^2 x} {\d t^2} + t \frac {\d x} {\d t} + \paren {t^2 - \alpha^2} x} s 
      | r = 0 
      | c = Laplace Transform of Bessel's Equation
}}
{{eqn | l = \frac {\d^2} {\d s^2} \laptrans {x' '} - \frac \d {\d s} \laptrans {x'} + \frac {\d^2} {\d s^2} \laptrans x - \alpha^2 \laptrans x
      | r = 0
      | c = setting the initial conditions as $\map x 0 = 1, \ \map {x'} 0 = 0$
}}
{{eqn | l = \paren {s^2 + 1} \LL' ' \set x + 3 s \LL' \set x + \paren {1 - \alpha^2} \laptrans x
      | r = 0
      | c = Make the following change of variable $u = \sqrt {s^2 + 1} \laptrans x$
}}
{{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 = \paren {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 \paren { \paren {u' \sqrt {s^2 + 1} }^2}'
      | r = \frac 1 2 \alpha^2 \paren {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 \map \ln {\sqrt {s^2 + 1} - s}
      | c = 
}}
{{eqn | l = u
      | r = \paren {\sqrt {s^2 + 1} - s}^\alpha
      | c = reverting the substitution
}}
{{eqn | l = \laptrans x
      | r = \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }
      | c = 
}}
{{end-eqn}}
Hence:
{{begin-eqn}}
{{eqn | l = \map {\laptrans {\map {J_\alpha} t} } s
      | r = \dfrac {\paren {\sqrt {s^2 + 1} - s}^\alpha} {\sqrt {s^2 + 1} }
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Sine of Root

:$\laptrans {\sin \sqrt t} = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }$
where $\laptrans f$ denotes the Laplace transform of the function $f$.

Proof:
{{begin-eqn}}
{{eqn | l = \sin \sqrt t
      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {\sqrt t}^{2 n + 1} } {\paren {2 n + 1}!}
      | c = {{Defof|Real Sine Function}}
}}
{{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {\paren {2 n + 1}!} t^{n + \frac 1 2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \laptrans {\sin \sqrt t}
      | r = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\map \Gamma {n + \frac 3 2} } {\paren {2 n + 1}! s^{n + \frac 3 2} }
      | c = Laplace Transform of Power, Linear Combination of Laplace Transforms
}}
{{eqn | r = \frac 1 {s^{3/2} } \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {n + \frac 1 2} \map \Gamma {n + \frac 1 2} } {\paren {2 n + 1}! s^n}
      | c = Gamma Difference Equation
}}
{{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {\paren {2 n + 1} \paren {2 n}!} {2^{2 n} n! \paren {2 n + 1}! s^n}
      | c = Gamma Function of Positive Half-Integer
}}
{{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n} {2^{2 n} n! s^n}
      | c = 
}}
{{eqn | r = \frac {\sqrt \pi} {2 s^{3/2} } \sum_{n \mathop = 0}^\infty \frac 1 {n!} \paren {-\frac 1 {2^2 s} }^n
      | c = 
}}
{{eqn | r = \dfrac {\sqrt \pi} {2 s^{3/2} } e^{-1/\paren {2^2 s} }
      | c = {{Defof|Exponential Function/Real|subdef = Power Series Expansion|Exponential Function}}
}}
{{eqn | r = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }
      | c = simplifying
}}
{{end-eqn}}
{{qed}}
Theorem: Laplace Transform of Sine of Root

:$\laptrans {\sin \sqrt t} = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }$
where $\laptrans f$ denotes the Laplace transform of the function $f$.

Proof:
Let $\map y t := \sin \sqrt t$.
Differentiating twice {{WRT|Differentiation}} $t$, we get:
:$(1): \quad 4 t y' ' + 2 y' ' + y = 0$
Let $\map Y s = \laptrans {\map t y}$ be the Laplace transform of $y$.
Then taking the Laplace transform of $(1)$:
{{begin-eqn}}
{{eqn | l = -4 \map {\dfrac \d {\d s} } {\laptrans {\map {y' '} t} } + 2 \laptrans {\map {y'} t} + \laptrans {\map y t}
      | r = 0
      | c = Derivative of Laplace Transform
}}
{{eqn | ll= \leadsto
      | l = -4 \map {\dfrac \d {\d s} } {s^2 \, \map Y s - s \, \map y 0 - \map {y'} 0} + 2 \paren {s \, \map Y s - \map y 0} + \map Y s
      | r = 0
      | c = Laplace Transform of Derivative, Laplace Transform of Second Derivative
}}
{{eqn | ll= \leadsto
      | l = 4 s^2 \, \map {Y'} s - \paren {6 s - 1} \map Y s
      | r = 0
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = \map Y s
      | r = \dfrac c {s^{3/2} } \, \map \exp {\dfrac 1 {4 s} }
      | c = solving the differential equation
}}
{{end-eqn}}
For small $t$, we have:
:$\sin \sqrt t \sim \sqrt t$
and:
:$\laptrans {\sqrt t} = \dfrac {\sqrt \pi} {2 s^{3/2} }$
For large $t$, we have:
:$Y \sim \dfrac c {2 s^{3/2} }$
Hence by comparison:
:$c = \dfrac {\sqrt \pi} 2$
Hence:
:$\laptrans {\sin \sqrt t} = \dfrac {\sqrt \pi} {2 s^{3/2} } \map \exp {-\dfrac 1 {4 s} }$
{{qed}}
Theorem: Laplace Transform of Cosine of Root over Root

:$\laptrans {\dfrac {\cos \sqrt t} {\sqrt t} } = \sqrt {\dfrac \pi s} \, \map \exp {-\dfrac 1 {4 s} }$
where $\laptrans f$ denotes the Laplace transform of the function $f$.

Proof:
Let $\map f t = \sin \sqrt t$.
Then:
{{begin-eqn}}
{{eqn | l = \map {f'} t
      | r = \dfrac {\cos \sqrt t} {2 \sqrt t}
      | c = 
}}
{{eqn | l = \map f 0
      | r = 0
      | c = 
}}
{{end-eqn}}
So:
{{begin-eqn}}
{{eqn | l = \laptrans {\map {f'} t}
      | r = \dfrac 1 2 \laptrans {\dfrac {\cos \sqrt t} {\sqrt t} }
      | c = 
}}
{{eqn | r = s \map F s - \map f 0
      | c = Laplace Transform of Derivative
}}
{{eqn | r = \dfrac {\sqrt \pi} {2 s^{1/2} } \map \exp {-\dfrac 1 {4 s} }
      | c = Laplace Transform of Sine of Root
}}
{{eqn | ll= \leadsto
      | l = \laptrans {\dfrac {\cos \sqrt t} {\sqrt t} }
      | r = \sqrt {\dfrac \pi s} \map \exp {-\dfrac 1 {4 s} }
      | c = 
}}
{{end-eqn}}
{{qed}}