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Today I was thinking in exotic integrals. Here $\{x\} $ denotes the fractional part function, and let the Airy function $\operatorname{Ai}(x)$ for real numbers $x$. Then I would like to know

Question. Is it possible to deduce a good approximation of $$\int_0^\infty \{x\}\operatorname{Ai}(x) \, dx \text{ ?}$$ Many thanks.

If there are no mistakes in my calculations, I can deduce using the definition of the fractional part function $$I:=\int_0^\infty\{x\}\operatorname{Ai}(x) \, dx = \sum_{k=1}^\infty \int_{k-1}^k(x-(k-1))\operatorname{Ai}(x)\,dx,$$ thus $$I=\int_0^\infty x\operatorname{Ai}(x)\,dx-\sum_{k=1}^\infty(k-1)\int_{k-1}^k \operatorname{Ai}(x)\,dx=\frac 1 {3^{1/3}\Gamma\left(\frac 1 3\right)}-\sum_{k=1}^\infty (k-1)\int_{k-1}^k\operatorname{Ai}(x)\,dx.$$

I don't know if is a good way to get such approximation of $I$.

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  • $\begingroup$ Many thanks @MichaelHardy $\endgroup$ – user243301 May 7 '17 at 20:58

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