How to do this integral:

$$ I = \int_{0}^{\infty} J_{0}(x)dx ?$$

  • $\begingroup$ ...with one line of computer algebra... result: $0$. $\endgroup$ – David G. Stork Jun 7 at 3:15

We have $$L\{J_0(x):p\}=\frac{1}{\sqrt{1+p^2}} . . . . . (1)$$ (from Laplace transformation)

By the definition of Laplace transformation, $$L\{J_0(x):p\}=\int_0^{\infty} J_0(x) e^{-px} dx$$

So from $(1)$, $$\int_0^{\infty} J_0(x) e^{-px} dx=\frac{1}{\sqrt{1+p^2}}$$

Putting $p=0$ we have, $$\int_0^{\infty} J_0(x) dx=\frac{1}{\sqrt{1+0}}=1$$


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