nemired9

2022-01-14

Proving that an ${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\int }_{k}^{k+1}\mathrm{sin}\left(\mathrm{exp}\left(x\right)\right)dxdk$ integral converges

ambarakaq8

Expert

$I={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\int }_{{e}^{k}}^{{e}^{k+1}}\frac{\mathrm{sin}z}{z}dzdk={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\left[Si\left({e}^{k+1}\right)-Si\left({e}^{k}\right)\right]dk={\int }_{0}^{+\mathrm{\infty }}\frac{Si\left(eu\right)-Si\left(u\right)}{u}du$
and we may invoke Frullani's integral to get $I=\frac{\pi }{2}$

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