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anonymous
 4 years ago
Metamath
\[\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x\]
anonymous
 4 years ago
Metamath \[\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x\]

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@mukushla can u PLZ explain me the symbols?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(J_0\) is the bessel function of order 0

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1looks like you are up to something!!

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1\[ J_\alpha(x) = \sum_{m=0}^\infty \frac{(1)^m}{m! \, \Gamma(m+\alpha+1)} {\left(\tfrac{1}{2}x\right)}^{2m+\alpha} \] \[ \int_0^\infty \sum_{m=0}^\infty \frac{(1)^m}{m! \, \Gamma(m+1)}\left( {1 \over 2} a \sqrt{ 1 + x^2}\right)^{2m} \\ \]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1349011854953:dw

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1this doesn't looks like converging ... let's try some other.

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1For a=1, the numerical value seems to be 0.540302

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x=\frac{\cos a}{a}\]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1how do do that man?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0using the formula\[\large J_\lambda(z)=\frac{1}{2\pi i}(\frac{z}{2})^{\lambda} \int_C t^{\lambda+1} e^{t\frac{z^2}{4t}} \text{d}t\]
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