Integrals of the form $\int\frac{1}{x^m(x-a)^{1/n}}\mathrm{d}x$

Question

Whilst doing some independent research into the derivatives of the Hurwitz Zeta function 1 I began by considering a more generalised version, namely

$$\hat{\zeta} (k, x; \alpha) = \sum_{k=0}^\infty \frac{1}{(k+x^{\alpha})^s}$$

I am not too sure of the terminology here, but the "extended" part using $\alpha$ differs from the usual Hurwitz Zeta function. The usual machinery to arrive at a Fourier expansion of this leads us to consider the comple coefficients as

\begin{eqnarray} c_j &= \lim_{a \rightarrow 0}\displaystyle \int_a^1 \zeta (k, t, ; \alpha)e^{2 \pi i j t} \mathrm{d}t\\ &=\lim_{a \rightarrow 0}\displaystyle \int_a^1 \sum_{k=0}^\infty \frac{e^{2 \pi i j t}}{(k+t^{\alpha})^s}\mathrm{d}t \end{eqnarray}

With the change of variables $u=k+t^{\alpha}$ we find

$$c_j = \frac{1}{\alpha}\lim_{a \rightarrow 0}\displaystyle \int_{k+a^{\alpha}}^k \sum_{k=0}^\infty \frac{e^{2 \pi i j t}}{u^s}\frac{1}{(u-k)^{\frac{1}{\alpha}}}\mathrm{d}t$$

So I have potentally an an issue around zero, and the limits can be simplified, considerations over $\Re (s)$ can also be implemented but I'll leave that until I have some idea how to integrate things such as ($\forall \, x \neq a$)

$$\displaystyle \int \frac{1}{x^n(x-a)^{\frac{1}{m}}}\mathrm{d}x$$

An obvious choice is to go down a trigonometric route, but I'm having no joy in the generalised case (with indices $(m, n)$ above) unless I pick easy cases such as $(m,n)=(2, 1)$.

Any steer is greatly appreciated, Thanks

Answer 1

Too long for a comment.

For sure, using Mathematica, I obtain the result already given by @MariuszIwaniuk in comments, namely $$I(m,n)=\displaystyle \int \frac{dx}{x^m (x-a)^{\frac{1}{n}}}=\frac{x^{1-m} (x-a)^{1-\frac{1}{n}} }{a (m-1)}\,\, _2F_1\left(1,-m-\frac{1}{n}+2;2-m;\frac{x}{a}\right)$$

However, assigning to $m$ a positive integer value returns as a result ComplexInfinity because an Infinite expression 1/0 encountered. This does not happen if $m$ is not an integer.

Considering the case where $m$ is a positive integer, we have $$I(m,n)=\frac{n\, (x-a)^{1-\frac{1}{n}}}{a^m(n-1)}\,\, _2F_1\left(m,\frac{n-1}{n};\frac{2n-1}{n};\frac{a-x}{a}\right)$$