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mukushla
Group Title
Metamath
\[\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x\]
 one year ago
 one year ago
mukushla Group Title
Metamath \[\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x\]
 one year ago
 one year ago

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sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
What is J_o
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.0
@mukushla can u PLZ explain me the symbols?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
\(J_0\) is the bessel function of order 0
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
looks like you are up to something!!
 one year ago

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

experimentX Group TitleBest ResponseYou've already chosen the best response.1
dw:1349011854953:dw
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
this doesn't looks like converging ... let's try some other.
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
For a=1, the numerical value seems to be 0.540302
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
\[\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x=\frac{\cos a}{a}\]
 one year ago

experimentX Group TitleBest ResponseYou've already chosen the best response.1
how do do that man?
 one year ago

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