Integration of LegendreP

Question

I am trying to integrate a product of 2 Legendre polynomials as follows:
Integrate[LegendreP[1, x] LegendreP[2l+1, x], {x, -1, 1}]

I get the result:
Sin[2 l [Pi]]/(l (3 + 2 l) [Pi])

which is always 0 for integer values of l, but shouldn't the result be 0 for $2l+1neq1$ and 2/3 for $2l+1=1$, since the LegendreP are orthogonal?

Andreas · Answer

Look at
Table[Sin[2 l [Pi]]/(l (3 + 2 l) [Pi]), {l, -2, 2}]

and then try
Limit[Sin[2 l [Pi]]/(l (3 + 2 l) [Pi]), l -> 0]

SHuisman · Answer

Note that sometimes a general result is returned. Think of:
Integrate[x^p,x]

returns:
x^(1+p)/(1+p)

However,
Integrate[x^-1, x]

gives:
Log[x]

But that is not represented by the general result…

cvgmt · Answer

Add Assumptions by hand or calculate the Limit
Integrate[LegendreP[1, x] LegendreP[2 l + 1, x], {x, -1, 1}, 
 Assumptions -> l == 0]
Limit[Integrate[LegendreP[1, x] LegendreP[2 l + 1, x], {x, -1, 1}], 
 l -> 0]
Integrate[LegendreP[1, x] LegendreP[2 l + 1, x], {x, -1, 1}, 
 Assumptions -> 3 + 2 l == 0]
Limit[Integrate[LegendreP[1, x] LegendreP[2 l + 1, x], {x, -1, 1}], 
 l -> -(3/2)]

Bob Hanlon · Answer

Clear["Global`*"]

Use Piecewise to handle the special case.
f[l_, x_] = 
 Piecewise[{{Integrate[LegendreP[1, x] LegendreP[2 l + 1, x], {x, -1, 1}], 
    ll != 0}, {Integrate[LegendreP[1, x] LegendreP[1, x], {x, -1, 1}], 
    l == 0}}]

Simplify[f[l, x], Element[l, Integers] && l != 0]

(* 0 *)

Integration of LegendreP

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