Reasoning with integrals of exponential functions

Question

I am going through the classic proof by Jitsuro Nagura

He presents the following as self-evident:

$$ \mbox{For}\ x \ge 1\ \mbox{and}\ n > 1:\quad \int_0^{\infty}\frac{1}{1 - \mathrm{e}^{-t}} \left[\mathrm{e}^{\large -\left(x + n - 1\right)\,t/n} - \mathrm{e}^{\large -xt}\right] \mathrm{d}t > 0 $$

It is not obvious to me that this is true. I wanted to understand the reasoning behind this.

Could someone explain high level why this is true for $x \geq 1$ and not true for $x < 1$.

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Edit:

Added $n>1$ based on feedback from Dmitry.

Answer 1

Note since $0 \lt e^{-t} \lt 1$ for $t \in (0,\infty)$, you have

$$\frac{1}{1 - e^{-t}} \gt 0 \tag{1}\label{eq1A}$$

Next, for $x \gt 1$ and $n \gt 1$, you get

$$\begin{equation}\begin{aligned} x & \gt 1 \\ x(n - 1) & \gt n - 1 \\ xn - x & \gt n - 1 \\ xn & \gt x + n - 1 \\ x & \gt \frac{x + n - 1}{n} \\ -x & \lt -\frac{x + n - 1}{n} \end{aligned}\end{equation}\tag{2}\label{eq2A}$$

This means for $t \gt 0$,

$$e^{-xt} \lt e^{-\frac{x + n - 1}{n}t} \implies e^{-\frac{x + n - 1}{n}t} - e^{-xt} \gt 0 \tag{3}\label{eq3A}$$

This shows both factors in the integrand of

$$\int_0^{\infty}\frac{1}{1-e^{-t}}\left(e^{-\frac{x+n-1}{n}t} - e^{-xt}\right)dt \tag{4}\label{eq4A}$$

for $t \gt 0$ are positive. Thus, if the integral converges, it must be positive.

To show it converges, the first issue to check on is the behavior of the integrand as $t \to 0$. Since it becomes $\frac{0}{0}$, you can use L'Hôpital's rule to get

$$\begin{equation}\begin{aligned} \lim_{t \to 0} \frac{e^{-\frac{x+n-1}{n}t} - e^{-xt}}{1-e^{-t}} & = \lim_{t \to 0} \frac{-\frac{x+n-1}{n}e^{-\frac{x+n-1}{n}t} - xe^{-xt}}{-e^{-t}} \\ & = \frac{x + n - 1}{n} + x \end{aligned}\end{equation}\tag{5}\label{eq5A}$$

This means the integral's lower limit is valid, so just treat the integral portion from $0$ to a fixed value, say $1$, as some positive constant value. Next, for $t \gt 1$, you have

$$\frac{1}{1 - e^{-t}} \lt \frac{1}{1 - e^{-1}} = j \tag{6}\label{eq6A}$$

so using this and that the exponential function is positive gives

$$0 \lt \int_1^{\infty}\frac{1}{1-e^{-t}}\left(e^{-\frac{x+n-1}{n}t} - e^{-xt}\right)dt \lt j\int_1^{\infty}e^{-\frac{x+n-1}{n}t}dt \tag{7}\label{eq7A}$$

For easier algebraic manipulations, have

$$k = \frac{x + n - 1}{n} \gt 0 \tag{8}\label{eq8A}$$

You thus get

$$\begin{equation}\begin{aligned} j\int_1^{\infty}e^{-kt}dt & = \left. j\left(\frac{e^{-kt}}{-k}\right)\right\rvert_{1}^{\infty} \\ & = \frac{je^{-k}}{k} \end{aligned}\end{equation}\tag{9}\label{eq9A}$$

Since this shows the middle integral in \eqref{eq7A} is bounded, it means that portion of the integral converges. Combined with the integral from $0$ to $1$ means the overall integral converges to a positive value.

Regarding the issue of the author using $x \ge 1$, note for $x = 1$ you have $-\frac{x+n-1}{n}t = -\frac{n}{n}t = -t$ and $-xt = -t$, so the integrand is always $0$ and, thus, the integral would always be $0$ (also, if $x \lt 1$, the value would be negative instead). There appears to be a typo, with the intent likely being $x \gt 1$, or that the integral is $\ge 0$ instead of being $\gt 0$. I haven't read the paper myself, so I'm not sure if one of these $2$ possibilities is the case, or if perhaps there's actually an error.

Answer 2

Even if it is not the most pleasant, the antiderivative exists $$I=\int\frac{e^{-\frac{x+n-1}{n}t} - e^{-xt}}{1-e^{-t}}\,dt$$ $$I=\frac{e^{t-t x} \, _2F_1\left(1,1-x;2-x;e^t\right)-n e^{\frac{t-t x}{n}} \, _2F_1\left(1,\frac{1-x}{n};\frac{n-x+1}{n};e^t\right)}{1-x}$$ $$J(x)=\int_0^\infty\frac{e^{-\frac{x+n-1}{n}t} - e^{-xt}}{1-e^{-t}}\,dt=\psi (x)-\psi \left(\frac{n+x-1}{n}\right)$$ provided that $\Re(x)>0\land \Re\left(\frac{1-x}{n}\right)<1$.

We have $\forall n$, $J(1)=0$ and this is the only root. Since $J(x)$ is an increasing function, then $J(x) >0$ if $x>1$ and $J(x) <0$ if $x<1$.

If you expand $J(x)$ around $x=1$, you have $$J(x)=\frac{\pi ^2 (n-1) }{6 n}(x-1)+O\left((x-1)^2\right)$$