I would be interested in finding a closed form or, at least, bounding (in terms of $m$ as it becomes larger) the real part of the following itnegral:
$$f(m,a):=\int_0^{2 \pi i m} \frac{e^{-t}}{t-a} dt$$
where $m \in \mathbb{N}, m>>0$ and $a \in \mathbb{C}, \mathfrak{Re}( a )>>0$.
For this purpose, I have tried three different methods.
Attempt 1: Firstly, I just tried to express the integral as a sum of integrals and made a change of variables, so that
$$f(m,a)= i \sum_{n=0}^{m-1} \int_{2 \pi n}^{2 \pi (n+1)} \frac{e^{-it}}{it-a} dt$$
Then, I compared those integrals whith the following:
$$\int_{\gamma} \frac{1}{\log{z}-a} dz $$
Where the path of integration is the unit circle clockwise. Each integral from sum of integrals correspond to different branches of the complex logarithm. To analyse this last one, I used a keyhole contour, taking the branch cut of the logarithm at the positive real axis. I obtained, for example, for the first summand (corresponding to the first branch of the logarithm),
$$\int_{0}^{2 \pi} \frac{e^{-it}}{it-a} dt = iR\int_0^{2 \pi} \frac{e^{it}}{2 \pi i - \log{R} -i t -a}dt - \int_1^R \frac{2 i \pi}{(a+\log{t})(a+\log{t}-2 i \pi)}dt$$
For $R \in \mathbb{R}, R > |a|$. My problem is that this family of integrals seems even more difficult to bound (even letting $R \to \infty$), so I have not gained anything.
Attempt 2: We could multiply and divide by $e^a$, so that we get:
$$f(m,a)=e^{-a} \int_0^{2 \pi i m} \frac{e^{-(t-a)}}{t-a} dt$$
From here, we could relate it to the Exponential Integral. However, the bounds that can be obtained this way seem to be not accurate at all for big $m$ and $\mathfrak{Re} a$, and it is difficult to extract the real part from them.
Attempt 3: We could separe the real and imaginary part of the original integral:
$$f(m,a)= -i \int_0^{2 \pi m} \frac{(\cos{t} + i \sin{t})(i\left (t + \mathfrak{Im}(a) \right) + \mathfrak{Re}(a))}{\mathfrak{Re}(a)^2+\left (\mathfrak{Im}(a) + t \right) ^2} dt$$
$$\mathfrak{Re}(f(m,a))= \int_0^{2 \pi m} \frac{\left (\mathfrak{Im}(a) + t \right) \cos{t} + \mathfrak{Re}(a) \sin{t}}{\mathfrak{Re}(a)^2+\left (\mathfrak{Im}(a) + t \right) ^2} dt$$
However, I do not know how to solve or bound this last integral neither.
Attempt 4: Using the following notation:
$$\mathscr{F}\left\{f(x)\right\} = F(s) = \int_{-\infty}^{\infty} {f(x)e^{-2\pi i sx} }dx$$ $$ \mathrm{sinc}(x) = \dfrac{\sin(\pi x)}{\pi x}$$ $$ \Pi(x) = \begin{cases} 1 \quad |x|<\frac{1}{2} \\ 0 \quad |x|>\frac{1}{2}\\ \end{cases}$$ $$ H(x) = \begin{cases} 0 \quad x < 0 \\ 1 \quad x > 0\\ \end{cases}$$
We can express our function as:
$$\int_{-\infty}^{\infty}\dfrac{1}{x - \dfrac{a}{2\pi i m}}\Pi\left(x-\frac{1}{2}\right)e^{-2\pi i mx}dx$$
Let's use the substitution $z_0 = \dfrac{a}{2\pi i m}$. Then, by making use of
$$\mathscr{F}\left\{\dfrac{1}{x-z_0}\right\} = -i\pi e^{-2\pi i s z_0}\left[\mathrm{sgn}(s)-\mathrm{sgn}\left(\Im\left[z_0\right]\right)\right]$$
We have:
$$\begin{align*} f(m,a) &= \mathscr{F}\left\{\dfrac{1}{x - z_0}\cdot\Pi\left(x-\frac{1}{2}\right)\right\} \\ &= \mathscr{F}\left\{\dfrac{1}{x - \dfrac{a}{2\pi i m}}\right\} * \mathscr{F}\left\{\Pi\left(x-\frac{1}{2}\right)\right\}\\ \\ &= -i\pi e^{-2\pi i m z_0}\left[\mathrm{sgn}(m)-\mathrm{sgn}\left(\mathfrak{Im}\left[z_0\right]\right)\right] * e^{-i\pi m}\mathrm{sinc}(m) \end{align*}$$
We can develop this a little bit more, but afer writing down the convolution as an integral we would end up with expressions that are pretty similar to the integral we are trying to solve, so that it would be useless.
This post has been cross-posted from MSE, where it received some upvotes but no definitive answer. I am curious about this problem, since the integral is seemingly inoffensive at first sight.
Any help will be welcomed.
Thank you.
