Integral representation of Bessel function $J_1(x)$

Solution 1:

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 $$\begin{align} - J_{1}(x) &= \frac{2}{\pi} \int_{1}^{\infty} \left(\frac{t}{\sqrt{t^{2}-1}} -1 \right) \cos(xt) \, \mathrm dt- \frac{2}{\pi} \frac{\sin (x)}{x} \\ &= \frac{2}{\pi} \int_{\color{red}{0}}^{\infty} \left(\frac{t \theta(t-1)}{\sqrt{t^{2}-1}} -1 \right) \cos(xt) \, \mathrm dt + \frac{2}{\pi} \int_{0}^{1} \cos(xt) \, \mathrm dt- \frac{2}{\pi} \frac{\sin (x)}{x}\\ &= \frac{2}{\pi} \int_{0}^{\infty} \left(\frac{t \theta(t-1)}{\sqrt{t^{2}-1}} -1 \right) \cos(xt) \, \mathrm dt. \end{align}$$