Laplace Transform of a Function which is related to the Bessel Function I am trying to prove the following:
Let $f(t)=\frac{J_{1}(kt)}{t}$ where $J_{1}$ is the Bessel function then show that the laplace transform of $f$ is $\frac {\sqrt {k^{2}+z^{2}}-z}{k}$.
I try to compute first the laplace transform of $J_{1}(t)$ by calculating the integral in the definition but I end it up with the integral $\frac{1}{\pi i} \int _{|w|=1} \frac{w}{iw^{2}+2zw-i} dw$ since $z$ can be any complex number I didnt get any where from this point. Any help would be great. Thanks.
 A: I found the LT by using the series representation of $J_1$.  Recall that
$$J_1(x) = \sum_{m=0}^{\infty} \frac{(-1)^m}{m! (m+1)!} \left (\frac{x}{2} \right )^{2 m+1}$$
Then
$$\begin{align}\int_0^{\infty} dt \frac{J_1(k t)}{t} e^{-s t} &= \sum_{m=0}^{\infty} \frac{(-1)^m}{m! (m+1)!} \int_0^{\infty} dt \frac1{t}\left (\frac{k t}{2} \right )^{2 m+1} e^{-s t} \\ &= \sum_{m=0}^{\infty} \frac{(-1)^m}{m! (m+1)!} \left (\frac{k}{2 s} \right )^{2 m+1} (2 m)! \\ &=\frac{s}{k} \sum_{m=0}^{\infty} \frac{(-1)^m}{2 m+2} \frac1{2^{2 m}} \binom{2 m}{m} \left (\frac{k}{s} \right )^{2 m+2} \end{align}$$
Let
$$f(x) = \sum_{m=0}^{\infty} \frac{(-1)^m}{2 m+2} \frac1{2^{2 m}} \binom{2 m}{m} x^{2 m+2} $$
Then
$$f'(x) = x \sum_{m=0}^{\infty} (-1)^m \frac1{2^{2 m}} \binom{2 m}{m} x^{2 m} = \frac{x}{\sqrt{1+x^2}}$$
$$\implies f(x) = \sqrt{1+x^2}+C$$
$f(0) = 0 \implies C=-1$.  Thus, the LT is
$$\int_0^{\infty} dt \frac{J_1(k t)}{t} e^{-s t} = \frac{\displaystyle \sqrt{1+\frac{k^2}{s^2}}-1}{k/s} $$
The result follows.
