Difference between revisions of "Relationship between Scorer Gi and Airy functions"
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| − | + | ==Theorem== | |
| − | + | The following formula holds: | |
| − | $$\mathrm{Gi}(x)=\mathrm{Bi}(x)\displaystyle\int_x^{\infty} \mathrm{Ai}(t) | + | $$\mathrm{Gi}(x)=\mathrm{Bi}(x)\displaystyle\int_x^{\infty} \mathrm{Ai}(t)\mathrm{d}t + \mathrm{Ai}(x)\displaystyle\int_0^x \mathrm{Bi}(t) \mathrm{d}t,$$ |
where $\mathrm{Gi}$ denotes the [[Scorer Gi]] function, $\mathrm{Ai}$ denotes the [[Airy Ai]] function, and $\mathrm{Bi}$ denotes the [[Airy Bi]] function. | where $\mathrm{Gi}$ denotes the [[Scorer Gi]] function, $\mathrm{Ai}$ denotes the [[Airy Ai]] function, and $\mathrm{Bi}$ denotes the [[Airy Bi]] function. | ||
| − | + | ||
| − | + | ==Proof== | |
| − | + | ||
| − | + | ==References== | |
| + | |||
| + | [[Category:Theorem]] | ||
| + | [[Category:Unproven]] | ||
Latest revision as of 22:53, 9 June 2016
Theorem
The following formula holds: $$\mathrm{Gi}(x)=\mathrm{Bi}(x)\displaystyle\int_x^{\infty} \mathrm{Ai}(t)\mathrm{d}t + \mathrm{Ai}(x)\displaystyle\int_0^x \mathrm{Bi}(t) \mathrm{d}t,$$ where $\mathrm{Gi}$ denotes the Scorer Gi function, $\mathrm{Ai}$ denotes the Airy Ai function, and $\mathrm{Bi}$ denotes the Airy Bi function.
