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mukushla

  • 2 years ago

Meta-math \[\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x\]

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  1. sauravshakya
    • 2 years ago
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    What is J_o

  2. sauravshakya
    • 2 years ago
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    @mukushla can u PLZ explain me the symbols?

  3. mukushla
    • 2 years ago
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    \(J_0\) is the bessel function of order 0

  4. experimentX
    • 2 years ago
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    looks like you are up to something!!

  5. experimentX
    • 2 years ago
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    \[ J_\alpha(x) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+\alpha+1)} {\left(\tfrac{1}{2}x\right)}^{2m+\alpha} \] \[ \int_0^\infty \sum_{m=0}^\infty \frac{(-1)^m}{m! \, \Gamma(m+1)}\left( {1 \over 2} a \sqrt{ 1 + x^2}\right)^{2m} \\ \]

  6. experimentX
    • 2 years ago
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    |dw:1349011854953:dw|

  7. experimentX
    • 2 years ago
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    this doesn't looks like converging ... let's try some other.

  8. experimentX
    • 2 years ago
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    For a=1, the numerical value seems to be 0.540302

  9. mukushla
    • 2 years ago
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    \[\int_{0}^{\infty} \text{J}_0(a\sqrt{1+x^2}) \ \text{d}x=\frac{\cos a}{a}\]

  10. experimentX
    • 2 years ago
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    how do do that man?

  11. mukushla
    • 2 years ago
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    using the formula\[\large J_\lambda(z)=\frac{1}{2\pi i}(\frac{z}{2})^{\lambda} \int_C t^{-\lambda+1} e^{t-\frac{z^2}{4t}} \text{d}t\]

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