Approximating $\displaystyle \int_{-\infty}^{\infty}\frac{e^{R+i\,u}}{\ln(R+i\,u)}du$ Working with a special function related with reciprocal gamma function, I've found the following integral
$$\int_{-\infty}^{\infty}\frac{e^{R+i\,u}}{(R+i\,u)\,\ln^2(R+i\,u)}du$$
with $R>1$. Integrating by parts we have :
$$\int_{-\infty}^{\infty}\frac{e^{R+i\,u}}{(R+i\,u)\,\ln^2(R+i\,u)}du=\left.\frac{i\,e^{R+i\,u}}{\ln(R+i\,u)}\right|_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\frac{e^{R+i\,u}}{\ln(R+i\,u)}du=$$
$$=\int_{-\infty}^{\infty}\frac{e^{R+i\,u}}{\ln(R+i\,u)}du$$
I wonder if there is any way to calculate any of them or at least approximate them. Thanks in advance. 
Progress: Using the Residue Theorem we can write the first integral as
$$\int_{-\infty}^{\infty}\frac{e^{R+i\,u}}{\ln(R+i\,u)}du=2\pi\, e+\int_{-\infty}^{\infty}\frac{e^{i\,u}}{\ln(i\,u)}du=$$
$$=2\pi\,e\,+\,2\int_0^{\infty}\frac{\cos(u)\ln(u)+\sin(u)\pi/2}{\ln^2(u)+(\pi/2)^2}du$$
 A: We can write the integral as:
$$I = -i\int_B\frac{\exp(z)}{\log(z)}dz$$
where $B$ is the Bromwich contour that moves from $R -i \infty $ to  $R +i \infty $. By Cauchy's theorem, we're free to move the contour to the left, but it must pass the singularities of the integrand to the right. We can then deform the contour such that it is to the left of the imaginary axis in the lower half plane below $-i \delta$ and then moves to the right to pass the pole at $z = 1$  to the right and then it moves back to the left of the imaginary axis to move to infinity parallel to the imaginary axis. The value of the integral is by Cauchy's theorem still the same.
To compute the integral we can complete the contour in the left half plane, but we must then deal with the branch singularity at $z = 0$. The logarithm can be defined by putting the branch cut along the negative real axis. We can then consider the contour consisting of the deformed Bromwich part, a quarter circle to just above the negative real axis, then from $-\infty + i\delta$ to $-\epsilon + i\delta$, then a clockwise circle with radius $\epsilon$ around the branch point at $z = 0$, and then we return along a line a distance $\delta$ below the negative real axis to $-\infty$. Finally we return to the starting point of the deformed Bromwich contour along a quarter circle.
If we want to be rigorous here we must, of course replace $\infty$ everwhere by some large $R$ and take the limit of $R$ to infinity. One can then show that the integrals along the quarter circles tend to zero, and one is left with the result:
$$-i\oint_C \frac{\exp(z)}{\log(z)}dz = I -2\pi \int_0^{\infty}\frac{\exp(-x)}{\log^2(x)+\pi^2}dx$$
By the residue theorem this contour integral is also equal to $2\pi i$ times the residue at $z = 1$, which yields $2\pi e$. Therefore, the integral can be written as:
$$I = 2\pi\left(e + \int_0^{\infty}\frac{\exp(-x)}{\log^2(x)+\pi^2}dx\right)$$
Numerically evaluating the integral in this expression is now easy as it converges fast:
$$I \approx 17.641740730649664316306264588647916877278\ldots$$ 
