Product of two generalized hypergeometric functions

Question

Is there a formula that allows me to calculate this product: $$\, _0F_1(;2;\text{ax}) \, _0F_1(;2;\text{bx})$$ Are there any references? Thanks in advance.

Answer 1

Yes, there is a general result, via which this can be cast in terms of the $_2F_1$ hypergeometric function. The relevant relation is formula 4.3.(12) on page 187 in "Higher Transcendental Functions, Vol. 1" by A. Erdelyi (Ed.), which I am reproducing below for your convenience.

$$ {}_{0}F_{1}(c;p z) \ {}_{0}F_{1}(c';q z)= \\ \sum_{n=0}^{\infty}\ \frac{(pz)^{n}}{n!\ (c)_{n}}\ \ {}_2F_{1}(1-c-n,-n; \ c'; \ \frac{q}{p}), $$

The symbol in the denominator is the usual Pochhammer.

The product you seek is the even more special case of this, where $c = c'$.

Answer 2

If you look here, you will see that this hypergeometric function is very closely related to the modified Bessel function of the first kind $$\, _0F_1(;2;a x)=\frac{I_1\left(2 \sqrt{a} \sqrt{x}\right)}{\sqrt{a} \sqrt{x}}$$ which makes $$\, _0F_1(;2;a x)\,\,\, _0F_1(;2;b x)=\frac{I_1\left(2 \sqrt{a} \sqrt{x}\right) I_1\left(2 \sqrt{b} \sqrt{x}\right)}{\sqrt{a} \sqrt{b}\, x}$$ I do not think that you could further simplify.