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Something is strange with $\sf LegendreQ$. Let $x>1$. I wonder why $\sf LegendreQ[\frac12,x]$ is complex-valued, and the following two codes do not give the same results:



(whose identity is given by Equation 8.820.2 of Gradshteyn & Ryzhik, 7th Ed.)

Thank you for your interest and time.

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Please post code rather than LaTeX. – MarcoB Feb 15 at 6:28
Also, you should not use the bugs tag until the behavior is confirmed to be a bug by the community or by Wolfram support. – MarcoB Feb 15 at 6:34
The code is posted. Thanks for the proper tag. – scieng Feb 15 at 9:48
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up vote 3 down vote accepted

I don't have access to the book right now, but it probably states

LegendreQ[1/2, 0, 3, x] == Pi/2/(2 x)^(3/2) Hypergeometric2F1[5/4, 3/4, 2, 1/x^2]
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I looked it up: the pre-factor works out to $\pi/(4 \sqrt{2})$ rather than $\pi/2$, but the rest holds as you wrote it. – MarcoB Feb 15 at 7:15
${\sf LegendreQ}[1/2,0,3,x]$ works. Thank you. By the way, the equality holds for $x>0$ according to $\it \text{}$. Somewhat confusing. – scieng Feb 15 at 11:57

As the documentation says, LegendreQ[n,m,a,z] gives Legendre functions of type a. The default is type 1. LegendreQ of types 1, 2, and 3 are defined in terms of LegendreP of these types, and have the same branch cut structure. Type 3 functions have a single branch cut from -Infinity to +1. So, as in the answer above, you need to specify type 3. And

Plot[{LegendreQ[1/2, 0, 3, x], 
    Pi/2/(2 x)^(3/2) Hypergeometric2F1[5/4, 3/4, 2, 1/x^2]},{x,1,5}]

looks fine.

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Beyond $x>1$, the equality holds for $x>0$ according to $\it \text{}$. This confuses me. – scieng Feb 15 at 11:59

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