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Question Number 146054 by mnjuly1970 last updated on 10/Jul/21

$$$$$$\mathrm{I}:={\int }_0^{\mathrm{\infty }}{e}^{-x}.{\mathrm{J}}_{\frac{1}{2}}\left(x\right)dx$$$${\mathrm{J}}_v\left(x\right)=x^v\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n{x}^{2n}}{2^{n+v}n!\mathrm{\Gamma }(n+v+1)}$$$$....$$

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