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Let $A>0$ and $0\le \mu \le 2$. Consider a following integral. \begin{equation} {\mathcal I}(A,\mu) := \int\limits_0^\infty e^{-(k A)^\mu} \cdot \frac{\cos(k)-1}{k} dk \end{equation} By substituting for $k A$ and then by expanding the cosine in a Taylor series about zero I have shown that: \begin{equation} {\mathcal I}(A,\mu) = \frac{1}{\mu} \sum\limits_{n=1}^\infty \frac{(1/A)^n}{n!} \cos\left(\frac{\pi}{2} n\right) \cdot \Gamma\left(\frac{n}{\mu}\right) \end{equation} Unfortunately the series on the right hand side above does not converge for small values of $A$. My question is therefore how do we find the small-$A$ behavior of ${\mathcal I}(A,\mu)$ ?

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Here we provide an answer for $\mu=2q/p$ where $p$ is a positive integer and $p+1\le 2q \le 2 p$. By using the multiplication theorem for the Gamma function we have shown that: \begin{eqnarray} &&{\mathcal I}(A,\mu) =\\ &&\frac{1}{\mu} \sqrt{\frac{(2\pi)^{2q-p}}{p 2q}} \sum\limits_{r=1}^q \left(\frac{p^{p/q}}{(2q)^2} \frac{(-1)}{A^2}\right)^r \cdot \frac{\prod\limits_{k=0}^{p-1} \Gamma(\frac{r}{q}+\frac{k}{p})}{\prod\limits_{k=0}^{2q-1} \Gamma(\frac{2r+1}{2q} + \frac{k}{2q})} \cdot F_{p+1,2q} \left[\begin{array}{rr} 1 & \left\{\frac{r}{q}+\frac{k}{p} \right\}_{k=0}^{p-1} \\ & \left\{\frac{2r+k}{2q}\right\}_{k=1}^{2q} \end{array}; \left(\frac{p^{p/q}}{(2q)^2} \frac{(-1)}{A^2}\right)^q\right] \end{eqnarray} Now we use the asymptotic behaviour of hypergeometric functions, see for example Wolfram's site. To make things simpler we plug the expression above into Mathematica and expand it using the Series[] command. The result is the following: \begin{eqnarray} {\mathcal I}(A,\frac{2}{1}) &=& \log(A) - \frac{1}{2} \gamma + A^2 + O(A^4)\\ {\mathcal I}(A,\frac{3}{2}) &=& \log(A) - \frac{1}{3} \gamma + A^{3/2} \frac{1}{2} \sqrt{\frac{\pi}{2}} + O\left((A^{3/2})^2\right) \\ {\mathcal I}(A,\frac{4}{3}) &=& \log(A) - \frac{1}{4} \gamma + A^{4/3} \frac{9 \sqrt{3} \pi }{280 \Gamma \left(-\frac{10}{3}\right)} + O\left((A^{4/3})^2\right) \end{eqnarray} As it seems the function in question is a linear combination of $\log(A)$ and some function that is analytic in $A^\mu$. It would be nice to find the full series expansion of that later function.

Since using Mathematica is cheating here we explicitely treat the case of $\mu=2$. In this case we have: \begin{eqnarray} {\mathcal I}(A,2) &=& -\frac{\, _2F_2\left(1,1;\frac{3}{2},2;-\frac{1}{4 A^2}\right)}{4 A^2} \\ &=& -\frac{1}{8 A^2} \int\limits_0^1 \int\limits_0^1 (1-t)^{-1/2} \exp\left(-\frac{t \cdot t_1}{4 A^2}\right) dt dt_1 \\ &=&-\frac{1}{2} \int\limits_0^{\frac{1}{4 A^2}} \left(1 - 4 A^2 t\right)^{-1/2} \left(\frac{1- e^{-t}}{t}\right) dt \end{eqnarray} Now we expand the first factor in the integrand in a Taylor series. Therefore we have: \begin{eqnarray} -2 {\mathcal I}(A,2) &=& \log \left(\frac{1}{4 A^2}\right)+\Gamma \left(0,\frac{1}{4 A^2}\right)+\gamma +\\ &&\sum\limits_{n=1}^\infty \binom{-1/2}{n} (-1)^n \left[\frac{1}{n} - (4 A^2)^n \gamma(n,\frac{1}{4 A^2})\right] \end{eqnarray} Now, since \begin{equation} \sum\limits_{n=1}^\infty \binom{-1/2}{n} \frac{(-1)^n}{n} = \log(4) \end{equation} we are getting the final answer: \begin{eqnarray} {\mathcal I}(A,2) = \log(A) - \frac{\gamma}{2} - \frac{1}{2} \Gamma(0,\frac{1}{4 A^2}) - \frac{1}{2} \sum\limits_{n=1}^\infty \binom{-1/2}{n} (-1)^n (4 A^2)^n \gamma(n,\frac{1}{4 A^2}) \end{eqnarray} Note that our conjecture was not quite true. Indeed our function (at least in the case $\mu=2$) is a linear combination of a natural logarithm, and a function that has a nontrivial Laurant expansion about the origin(ie has all possible both positive and negative powers in the expansion).

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Clearly we have: \begin{equation} {\mathcal I}(A,\mu) = \underbrace{\int\limits_0^{1/A} e^{-(k A)^\mu} \left(\frac{\cos(k)-1}{k}\right) dk}_{{\mathcal T}_1}+ \underbrace{\int\limits_{1/A}^\infty e^{-(k A)^\mu} \left(\frac{\cos(k)-1}{k}\right) dk}_{{\mathcal T}_2} \end{equation} Now we analyze both terms. \begin{equation} {\mathcal T}_1 = \underbrace{\int_0^{1/A} \frac{\cos(k)-1}{k} dk}_{-\gamma+Ci(\frac{1}{A}) - \log(\frac{1}{A})} + \sum\limits_{n=1}^\infty \frac{(-A^\mu)^n}{n!} \underbrace{\int\limits_0^{1/A} k^{n \mu-1} \left(\cos(k)-1\right) d k}_{Re\left[\imath^{n\mu} \gamma(n \mu,-\frac{\imath}{A})\right]- \frac{1}{n\mu} \left(\frac{1}{A}\right)^{n\mu}} \end{equation} The first result on the right hand side above follows from: \begin{equation} \lim_{M \rightarrow \infty} \int\limits_0^M \frac{\cos(k)-1}{k} dk + \log(M) = -\gamma \end{equation} and the rest is straightforward. Therefore the first term equals: \begin{equation} {\mathcal T}_1 = -\gamma + Ci(\frac{1}{A}) - \log(\frac{1}{A}) + Re\left[\sum\limits_{n=1}^\infty \frac{(-1)^n (\imath A)^{n \mu}}{n!} \gamma(n \mu,-\frac{\imath}{A})\right] - \left(-\gamma+ Ei(-1)\right) \frac{1}{\mu} \end{equation} Now let us move on to the second term: \begin{equation} {\mathcal T}_2 = \int\limits_1^\infty e^{-k^\mu} \frac{\cos(\frac{k}{A})}{k} dk + \frac{1}{\mu} Ei(-1) \end{equation} In the integral above we expand the exponential in a Maclaurin series and integrate term by term. This gives: \begin{equation} {\mathcal T}_2 = Re\left[\Gamma(0,-\frac{\imath}{A})\right] + Re\left[\sum\limits_{n=1}^\infty \frac{(-1)^n (\imath A)^{n \mu}}{n!} \Gamma(n \mu,-\frac{\imath}{A})\right] + \frac{1}{\mu} Ei(-1) \end{equation} Combining both terms together we get a very simple result: \begin{equation} {\mathcal I}(A,\mu) = \log(A) - \gamma\left(1-\frac{1}{\mu}\right) + Re\sum\limits_{n=1}^\infty \frac{(-(\imath A)^\mu)^n}{n!} \Gamma(n \mu) \end{equation} Both the constant $Ei(-1)$ and the cosine integral function have miraculously canceled each other! Note: The series on the right hand side of the last equation converges only if $0<\mu<1$. If $1\le \mu \le 2$ this is an asymptotic series.

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