Limit behavior of a definite integral that depends on a parameter. 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 ${\mathcal I}(A,\mu)$ ?

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.
