Here's the question you clicked on:
mukushla
Meta-math \[\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x\]
@mukushla can u PLZ explain me the symbols?
\(J_0\) is the bessel function of order 0
looks like you are up to something!!
\[ J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\alpha+1)} {\left(\tfrac{1}{2}x\right)}^{2m+\alpha} \] \[ \int_0^\infty \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+1)}\left( {1 \over 2} a \sqrt{ 1 + x^2}\right)^{2m} \\ \]
|dw:1349011854953:dw|
this doesn't looks like converging ... let's try some other.
For a=1, the numerical value seems to be 0.540302
\[\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x=\frac{\cos a}{a}\]
how do do that man?
using the formula\[\large J_\lambda(z)=\frac{1}{2\pi i}(\frac{z}{2})^{\lambda} \int_C t^{-\lambda+1} e^{t-\frac{z^2}{4t}} \text{d}t\]