## mukushla 3 years ago Meta-math $\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x$

1. sauravshakya

What is J_o

2. sauravshakya

@mukushla can u PLZ explain me the symbols?

3. mukushla

$$J_0$$ is the bessel function of order 0

4. experimentX

looks like you are up to something!!

5. experimentX

$J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\alpha+1)} {\left(\tfrac{1}{2}x\right)}^{2m+\alpha}$ $\int_0^\infty \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+1)}\left( {1 \over 2} a \sqrt{ 1 + x^2}\right)^{2m} \\$

6. experimentX

|dw:1349011854953:dw|

7. experimentX

this doesn't looks like converging ... let's try some other.

8. experimentX

For a=1, the numerical value seems to be 0.540302

9. mukushla

$\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x=\frac{\cos a}{a}$

10. experimentX

how do do that man?

11. mukushla

using the formula$\large J_\lambda(z)=\frac{1}{2\pi i}(\frac{z}{2})^{\lambda} \int_C t^{-\lambda+1} e^{t-\frac{z^2}{4t}} \text{d}t$