An integral $\int_0^\infty P_s(x-1)\,e^{-x}\,dx$ involving Legendre functions

Question

Let $P_s(x)$ denote the Legendre functions of the $1^{st}$ kind, i.e. the Legendre polynomial generalized to an arbitrary (not necessarily integer) order $s$. It can be expressed using the hypergeometric function: $$P_s(x)={_2F_1}\left(-s,s+1;\ 1;\ \frac{1-x}2\right).\tag1$$ Let's consider the definite integral $$\mathcal{J}(s)=\int_0^\infty P_s(x-1)\,e^{-x}\,dx.\tag2$$ It evaluates to integer values when $s\in\mathbb{Z}^+$. Using a computer-assisted search for the general term formula, I discovered the following conjecture, that I haven't yet been able to prove: $$\mathcal{J}(s)\stackrel?=\left(K_{3/2}(1)\cdot I_{s+1/2}(-1)-I_{3/2}(-1)\cdot K_{s+1/2}(1)\right)\sqrt{-1},\tag3$$ where $I_\nu(z)$ and $K_\nu(z)$ are the modified Bessel functions of the $1^{st}$ and $2^{nd}$ kind.

Unfortunately, it only seems to hold for $s\in\mathbb{Z}^+$, and does not generalize to non-integer values of $s$.

• Can we prove the conjecture $(3)$?

• Can we find a more general formula that holds not only for integer values of $s$?

• Can we find (or at least conjecture) a closed form for $\mathcal{J}(1/2)$?

Answer 1

Not a proper answer, but it is the closest one to your integral that I was able to find. Using formula 7.141.1 in Gradshteyn-Ryzhik, after some simplifications, one can get: $$\int_0^\infty P_s(x+1)\,e^{-x}\,dx=\frac{e\,\sqrt2}{\sqrt\pi} K_{s+\frac{1}{2}}(1),$$ where $K_\nu(x)$ is the modified Bessel function of the 2nd kind. Note that this formula contains $(x+1)$ rather than $(x-1)$ that appears in your question. I'm still trying to find a proper answer...

Answer 2

For $s=0,1,2,\ldots$ we compute $$ \mathcal{J}(s) = 1, 0, 1, 5, 36, 329, 3655, 47844, 721315, 12310199, 234615096, \ldots $$ [This was obtained with the gp code

F(p) = sum(k=0, poldegree(p), k!*polcoeff(p,k))
N = 10
vector(N+1,s,F(pollegendre(s-1,x-1)))

and can also be obtained from the expansion of $P_s(x-1)$ in powers of $x$, giving $(-1)^s \sum_{k=0}^s 2^{-k} k! {s \choose k} {-s-1 \choose k}$.] This matches OEIS sequence 806 up to sign, indicating that we're dealing with $(-1)^s y_s(-1)$ where $y_s$ is the Bessel polynomial of degree $s$. There's a lot of information about these numbers in that OEIS entry, including a recurrence equivalent to $$ {\mathcal J}(s) = (2s-1) {\mathcal J}(s-1) + {\mathcal J}(s-2) $$ and a simpler formula in terms of modified Bessel functions:

a(n) = BesselK[n+1/2,-1] / BesselK[5/2,-1]

contributed by Vaclav Kotesovec on Aug 07 2013. These formulas, together with known recurrences for modified Bessel functions, should soon also yield Vladimir Reshetnikov's conjectural formula in terms of $I_{s+1/2}(-1)$ and $K_{s+1/2}(1)$.

Answer 3

$\int_0^\infty P_s(x-1)e^{-x}~dx$

$=\int_{-2}^\infty P_s(x+1)e^{-x-2}~dx$

$=\int_{-2}^0P_s(x+1)e^{-x-2}~dx+\int_0^\infty P_s(x+1)e^{-x-2}~dx$

$=\int_{-2}^0{_2F_1}\left(-s,s+1;1;-\dfrac{x}{2}\right)e^{-x-2}~dx+\dfrac{\sqrt2}{e\sqrt\pi}K_{s+\frac{1}{2}}(1)$ (according to OlegK's answer)

$=2e^{-2}\int_0^1{_2F_1}(-s,s+1;1;x)e^{2x}~dx+\dfrac{\sqrt2}{e\sqrt\pi}K_{s+\frac{1}{2}}(1)$

$=2e^{-2}\int_0^1\sum\limits_{n=0}^\infty\dfrac{2^nx^n{_2F_1}(-s,s+1;1;x)}{n!}dx+\dfrac{\sqrt2}{e\sqrt\pi}K_{s+\frac{1}{2}}(1)$

$=2e^{-2}\left[\sum\limits_{n=0}^\infty\dfrac{2^nx^{n+1}{_3F_2}(-s,s+1,n+1;1,n+2;x)}{n!(n+1)}\right]_0^1+\dfrac{\sqrt2}{e\sqrt\pi}K_{s+\frac{1}{2}}(1)$ (according to http://functions.wolfram.com/HypergeometricFunctions/Hypergeometric2F1/21/01/02/01)

$=\sum\limits_{n=0}^\infty\dfrac{2^{n+1}{_3F_2}(-s,s+1,n+1;1,n+2;1)}{(n+1)!e^2}+\dfrac{\sqrt2}{e\sqrt\pi}K_{s+\frac{1}{2}}(1)$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-s)_k(s+1)_k(n+1)_k2^{n+1}}{(n+1)!(1)_k(n+2)_kk!e^2}+\dfrac{\sqrt2}{e\sqrt\pi}K_{s+\frac{1}{2}}(1)$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(-s)_k(s+1)_k\Gamma(n+k+1)2^{n+1}}{n!(1)_k\Gamma(n+k+2)k!e^2}+\dfrac{\sqrt2}{e\sqrt\pi}K_{s+\frac{1}{2}}(1)$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^\infty\dfrac{(1)_{n+k}(-s)_k(s+1)_k2^{n+1}}{(2)_{n+k}n!(1)_kk!e^2}+\dfrac{\sqrt2}{e\sqrt\pi}K_{s+\frac{1}{2}}(1)$

$=2e^{-2}\mathrm{F}^{1:0;2}_{1:0;1}\Bigg(\begin{matrix}1&:&-&;&-s,s+1&\\2&:&-&;&1&\end{matrix}\Bigg|2,1\Bigg)+\dfrac{\sqrt2}{e\sqrt\pi}K_{s+\frac{1}{2}}(1)$