Solving or bounding the real part of the integral $\int_0^{2 \pi i m} \frac{e^{-t}}{t-a} dt$

Question

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 Fourier Transform method mentiones in the answer, we can work around with the integrals, but would end up with Exponential Integral (attempt 2).

Any help will be welcomed.

Thank you.

Answer 1

I can't solve or bound the integral for finite $m$, but I can give you a solution for $m \to \infty$ using Fourier Transforms.

Some notation conventions I'll use

$$\mathscr{F}\left\{f(x)\right\} = F(s) = \int_{-\infty}^{\infty} {f(x)e^{-2\pi i sx} }dx$$ $$ \mathrm{sinc}(x) = \dfrac{\sin(\pi x)}{\pi x}$$ $$ \Pi(x) = \begin{cases} 1 \quad |x|<\frac{1}{2} \\ 0 \quad |x|>\frac{1}{2}\\ \end{cases}$$ $$ H(x) = \begin{cases} 0 \quad x < 0 \\ 1 \quad x > 0\\ \end{cases}$$

I'll also use this Fourier Transform pair

$$\mathscr{F}\left\{\dfrac{1}{x-z_0}\right\} = -i\pi e^{-2\pi i s z_0}\left[\mathrm{sgn}(s)-\mathrm{sgn}\left(\Im\left[z_0\right]\right)\right]$$

Starting from your integral

$$\begin{align*} f(m,a) & =\int_0^{2 \pi i m} \frac{e^{-t}}{t-a} dt\\ \\ &= 2\pi i m \int_0^1 {\dfrac{e^{-2\pi i mx}}{2\pi i mx - a}}dx\\ \\ &= \int_0^1{\dfrac{e^{-2\pi i mx}}{x - \dfrac{a}{2\pi i m}}}dx\\ \\ &= \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\\ \\ &= \mathscr{F}\left\{\dfrac{1}{x - \dfrac{a}{2\pi i m}}\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 \frac{a}{2\pi i m}}\left[\mathrm{sgn}(m)-\mathrm{sgn}\left(\Im\left[\dfrac{a}{2\pi i m}\right]\right)\right] * e^{-i\pi m}\mathrm{sinc}(m)\\ \\ &= -i\pi e^{-a}\left[1-\mathrm{sgn}\left(\Im\left[\dfrac{-ia}{2\pi}\right]\right)\right]\mathrm{sgn}(m) * e^{-i\pi m}\mathrm{sinc}(m)\\ \\ &= -i\pi e^{-a}\left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\mathrm{sgn}(m) * e^{-i\pi m}\mathrm{sinc}(m)\\ \\ &= -i\pi e^{-a}\left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\left[2H(m)-1\right] * e^{-i\pi m}\mathrm{sinc}(m)\\ \\ &= -i\pi e^{-a}\left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\int_{-\infty}^\infty \left[2H(m-\tau)-1\right] e^{-i\pi \tau}\mathrm{sinc}(\tau)\space d\tau \\ \\ &= -i\pi e^{-a}\left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\left[2\int_{-\infty}^\infty H(m-\tau) e^{-i\pi \tau}\mathrm{sinc}(\tau)\space d\tau - \int_{-\infty}^\infty e^{-i\pi \tau}\mathrm{sinc}(\tau)\space d\tau\right] \\ \\ &= -i\pi e^{-a}\left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\left[2\int_{-\infty}^m e^{-i\pi \tau}\mathrm{sinc}(\tau)\space d\tau - \int_{-\infty}^\infty e^{-2\pi is \tau}\mathrm{sinc}(\tau)\space d\tau\biggr|_{s=\frac{1}{2}}\right] \\ \\ &= -i\pi e^{-a}\left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\left[2\int_{-\infty}^m e^{-i\pi \tau}\mathrm{sinc}(\tau)\space d\tau - \Pi(s)\biggr|_{s=\frac{1}{2}}\right] \\ \\ &= -i\pi e^{-a}\left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\left[2\int_{-\infty}^m e^{-i\pi \tau}\mathrm{sinc}(\tau)\space d\tau - \dfrac{\Pi\left({\frac{1}{2}}^-\right)+ \Pi\left({\frac{1}{2}}^+\right)}{2}\right] \\ \\ &= -i\pi e^{-\Re(a)}e^{-i\Im(a)}\left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\left[2\int_{-\infty}^m e^{-i\pi \tau}\mathrm{sinc}(\tau)\space d\tau - \dfrac{1}{2}\right] \\ \\ &= \left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\pi e^{-\Re(a)}\left[-\sin\left(\Im[a]\right)-i\cos\left(\Im[a]\right)\right]\left[2\int_{-\infty}^m e^{-i\pi \tau}\mathrm{sinc}(\tau)\space d\tau - \dfrac{1}{2}\right] \\ \\ \end{align*}$$

If $\Re(a) < 0$, then $f(m,a) = 0$.

Also $\lim_{\Re(a) \to \infty} f(m,a) = 0$.

For $\Re(a) \ge 0$, as $m \to \infty$

$$\begin{align*}\lim_{m \to \infty} f(m,a) &= \lim_{m \to \infty} \left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\pi e^{-\Re(a)}\left[-\sin\left(\Im[a]\right)-i\cos\left(\Im[a]\right)\right]\left[2\int_{-\infty}^m e^{-i\pi \tau}\mathrm{sinc}(\tau)\space d\tau - \dfrac{1}{2}\right] \\ \\ &= \left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\pi e^{-\Re(a)}\left[-\sin\left(\Im[a]\right)-i\cos\left(\Im[a]\right)\right]\left[2\int_{-\infty}^\infty e^{-i\pi \tau}\mathrm{sinc}(\tau)\space d\tau - \dfrac{1}{2}\right] \\ \\ &= \left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]\pi e^{-\Re(a)}\left[-\sin\left(\Im[a]\right)-i\cos\left(\Im[a]\right)\right]\left[2\Pi(s)|_{s=\frac{1}{2}} - \dfrac{1}{2}\right] \\ \\ &= \dfrac{\left[1+\mathrm{sgn}\left(\Re\left[a\right]\right)\right]}{2}\pi e^{-\Re(a)}\left[-\sin\left(\Im[a]\right)-i\cos\left(\Im[a]\right)\right] \\ \\ \end{align*}$$

which should make obvious the value to which $\Re[f(m,a)]$ is converging.

Unfortunately, the integral

$$\int_{-\infty}^m e^{-i\pi \tau}\mathrm{sinc}(\tau)\space d\tau$$

is problematic for finite $m$. The real part converges rapidly to $\frac{1}{2}$. However, I suspect its imaginary part diverges because $\Im\left[\frac{i \log(-\infty)}{2\pi}\right] \to \infty$. So I don't know how to compute it for finite $m$.

Answer 2

Assume that $a$ is fixed and has a positive real part. Take the incomplete gamma function $\Gamma(0, z)$ to have the branch cut along the negative real axis and to be continuous from above on the branch cut. Then $$\int_0^{2 \pi i m} \frac {e^{-t}} {t - a} dt = e^{-a} \int_{-a}^{-a + 2 \pi i m} \frac {e^{-t}} t dt = \\ e^{-a} (\Gamma(0, -a) - \Gamma(0, -a + 2 \pi i m) - 2 \pi i \,[\operatorname{Im} a > 0]),$$ since the antiderivative $-\Gamma(0, t)$ has a $2 \pi i$ jump if the segment $[-a, -a + 2 \pi i m]$ crosses the branch cut.

Then, from known asymptotics of $\Gamma(0, z)$, $$\int_0^{2 \pi i m} \frac {e^{-t}} {t - a} dt = e^{-a} (\Gamma(0, -a) - 2 \pi i \,[\operatorname{Im} a > 0]) - \frac 1 {2 \pi i m} + O {\left( \frac 1 {m^2} \right)}, \\ m \to \infty, \; m \in \mathbb N.$$ The real part will not contain the $1/(2 \pi i m)$ term.