Difference between revisions of "Relationship between Scorer Gi and Airy functions"

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===Theorem===
+
==Theorem==
 
The following formula holds:
 
The following formula holds:
$$\mathrm{Gi}(x)=\mathrm{Bi}(x)\displaystyle\int_x^{\infty} \mathrm{Ai}(t)dt + \mathrm{Ai}(x)\displaystyle\int_0^x \mathrm{Bi}(t)dt,$$
+
$$\mathrm{Gi}(x)=\mathrm{Bi}(x)\displaystyle\int_x^{\infty} \mathrm{Ai}(t)\mathrm{d}t + \mathrm{Ai}(x)\displaystyle\int_0^x \mathrm{Bi}(t) \mathrm{d}t,$$
 
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===
+
==Proof==
 +
 
 +
==References==
  
 
[[Category:Theorem]]
 
[[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.

Proof

References