<s>[INST] Integral representation of Bessel function $J_1(x)$ In "The Handbook of Mathematical Functions" by Abramovitz and Stegun, according to Eq. 9.1.24,
\begin{align}
J_0(x)=&\frac{2}{\pi}\int_{1}^\infty \frac{\sin(xt)}{\sqrt{t^2-1}}dt,\quad x>0.
\end{align}
Naively using $J_1(x)=-dJ_0(x)/dx$,
\begin{align}
J_1(x)=&-\frac{2}{\pi}\int_{1}^\infty  \frac{t\cos(xt)}{\sqrt{t^2-1}}dt,\quad x>0.
\end{align}
Neither one of these two integrals seem absolutely convergent, with the latter being particularly bad as the non-oscillatory part of the integrand becomes a constant for large $t$.
However, on feeding the left-hand-side of the following expression to WolfraAlpha, I obtain $J_1(x)$:
\begin{align}
-\frac{2}{\pi}\int_{0}^\infty  \bigg[\frac{t\Theta(t-1)}{\sqrt{t^2-1}}-1\bigg]\cos(xt)dt= J_1(x),\quad x>0.
\end{align}
($\Theta(x)$ is the Heaviside step function.) I am wondering how to analytically obtain this result.
 [/INST] If $x>0$, then
$$ \begin{align}J_{0}(x) &= \frac{2}{\pi} \int_{1}^{\infty} \frac{\sin (xt)}{\sqrt{t^{2}-1}} \, \mathrm dt \\ &= \frac{2}{\pi} \int_{1}^{\infty} \left(\frac{1}{\sqrt{t^{2}-1}} -\frac{1}{t} \right) \sin(xt) \, \mathrm dt + \frac{2}{\pi} \int_{1}^{\infty}\frac{\sin (xt)}{t} \, \mathrm dt \\ &= \frac{2}{\pi} \int_{1}^{\infty} \left(\frac{1}{\sqrt{t^{2}-1}} -\frac{1}{t} \right) \sin(xt) \, \mathrm dt+ \frac{2}{\pi} \int_{x}^{\infty} \frac{\sin (u)}{u} \, \mathrm du \\ &=  \frac{2}{\pi} \int_{1}^{\infty} \left(\frac{1}{\sqrt{t^{2}-1}} -\frac{1}{t} \right) \sin(xt) \, \mathrm dt + \frac{2}{\pi} \left(\frac{\pi}{2}- \operatorname{Si}(x) \right). \end{align}$$
Differentiating both sides of the equation, we get $$\begin{align} - J_{1}(x) &= \frac{2}{\pi} \int_{1}^{\infty} \left(\frac{t}{\sqrt{t^{2}-1}} -1  \right) \cos(xt) \, \mathrm dt- \frac{2}{\pi} \frac{\sin (x)}{x} \\ &=  \frac{2}{\pi} \int_{\color{red}{0}}^{\infty} \left(\frac{t \theta(t-1)}{\sqrt{t^{2}-1}} -1  \right) \cos(xt) \, \mathrm dt + \frac{2}{\pi} \int_{0}^{1} \cos(xt) \, \mathrm dt- \frac{2}{\pi} \frac{\sin (x)}{x}\\ &= \frac{2}{\pi} \int_{0}^{\infty} \left(\frac{t \theta(t-1)}{\sqrt{t^{2}-1}} -1  \right) \cos(xt) \, \mathrm dt. \end{align}$$
</s>