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I am sorry if the answer to my question is well-known. I am quite new in this topic, so it will be also nice to have a reference, if it exists.

I was wondered if there exists a nice closed formula for a logarithm of an arbitrary hypergeometric series in term the terms of, say, a linear combination of some other hypergeometric series. The reason that makes me believe in the existence of such a formula is the following.

It is an easy exercise to show that the derivative of a hypergeometric series can be expressed as follows: $\frac{d}{dx} {}_nF_m (a_1,\ldots, a_n;b_1\ldots b_m; x) = \frac {a_1\cdots a_n}{b_1\cdots b_m} {}_nF_m (a_1+1,\ldots, a_n+1;b_1+1\ldots b_m+1; x)$. From the other hand, for an arbitrary function $G(x)$ we have $(\log G(x))' = \frac {G'(x)}{G(x)}$.

It follows that it suffices to find a ratio of two hypergeometric series to find a logarithmic derivative. In some cases this ratio is known to be a hypergeometric series again. So after the integration we'll obtain the desired result.

Thank you in advance for any help.

1

Logarithm of a hypergeometric series

I am sorry if the answer to my question is well-known. I am quite new in this topic, so it will be also nice to have a reference, if it exists.

I was wondered if there exists a nice closed formula for a logarithm of an arbitrary hypergeometric series in term of, say, a linear combination of some other hypergeometric series. The reason that makes me believe in the existence of such a formula is the following.

It is an easy exercise to show that the derivative of a hypergeometric series can be expressed as follows: $\frac{d}{dx} {}_nF_m (a_1,\ldots, a_n;b_1\ldots b_m; x) = \frac {a_1\cdots a_n}{b_1\cdots b_m} {}_nF_m (a_1+1,\ldots, a_n+1;b_1+1\ldots b_m+1; x)$. From the other hand, for an arbitrary function $G(x)$ we have $(\log G(x))' = \frac {G'(x)}{G(x)}$.

It follows that it suffices to find a ratio of two hypergeometric series to find a logarithmic derivative. In some cases this ratio is known to be a hypergeometric series again. So after the integration we'll obtain the desired result.

Thank you in advance for any help.
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