Behaviour near a singular point

[akotronis]

Evaluate \(\displaystyle\lim_{x\to-e^+}\int_{0}^{+\infty}\frac{(x+e)^{1/2}}{e^t+xt}\,dt\), if it exists.

[r9m]

Making the change in variable $x \mapsto -ex$, the problem is equivalent to the determination of: \begin{align*} \lim_{x\to-e^+}\int_{0}^{+\infty}\frac{(x+e)^{1/2}}{e^t+xt}\,dt = \lim_{x \to 1^{-}} \frac{(1-x)^{1/2}}{\sqrt{e}}\int_{0}^{\infty}\frac{1}{e^{t-1} - xt}\,dt\end{align*}

Let, $x \in (0,1)$, then $\displaystyle e^{t-1} > t \implies \left|\frac{xt}{e^{t-1}}\right| < 1$ for $t \in \mathbb{R}^{+}$, thus \begin{align*} I(x) = \int_0^{\infty} \frac{1}{e^{t-1} - xt}\,dt &= \sum\limits_{n \ge 1} x^{n-1}e^{n}\int_0^{\infty} t^{n-1}e^{-nt}\,dt\\&= \sum\limits_{n \ge 1} \frac{e^n n! x^{n-1}}{n^{n+1}}
\end{align*}

Now, we recall Stirling's approximation, $\displaystyle \frac{e^{n} n!}{n^{n+1}} = \sqrt{\frac{2\pi}{n}}\left(1+\frac{1}{12n}+O\left(\frac{1}{n^2}\right)\right)$,

Hence, for $0 < x < 1$ we have, $$I(x) - \sqrt{2\pi}\left(\sum\limits_{n \ge 1} \frac{x^{n-1}}{\sqrt{n}}\right) = \sum\limits_{n \ge 1} \left(\frac{e^n n!}{n^{n+1}} - \sqrt{\frac{2\pi}{n}}\right)x^{n-1} = O\left(\sum\limits_{n \ge 1} \frac{x^{n-1}}{n^{3/2}}\right)$$ which is bounded. Thus, it suffices to determine the limit, $\displaystyle \lim\limits_{x \to 1^{-}} \sqrt{1-x}\left(\sum\limits_{n \ge 1} \frac{x^{n-1}}{\sqrt{n}}\right)$.

Now, $\displaystyle \left(\sum\limits_{n \ge 1} \frac{x^{n-1}}{\sqrt{n}}\right)^2 = \sum\limits_{n \ge 2} \left(\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{k(n-k)}}\right)x^{n-2} = \sum\limits_{n \ge 2} \left(\frac{1}{n}\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{\frac{k}{n}\left(1-\frac{k}{n}\right)}}\right)x^{n-2}$

The function $\displaystyle f(t) = \frac{1}{\sqrt{t(1-t)}}$ is convex in $(0,1)$, and hence, \begin{align*}&\int_{\frac{1}{n}}^{1 - \frac{1}{n}} f(t)\,dt + \frac{1}{2n}\left(f\left(\frac{1}{n}\right) + f\left(1-\frac{1}{n}\right)\right) \le \frac{1}{n}\sum\limits_{k=1}^{n-1} f\left(\frac{k}{n}\right) < \int_0^{1} f(t)\,dt \\ \implies & \pi - 4\arcsin \frac{1}{\sqrt{n}} < \frac{1}{n}\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{\frac{k}{n}\left(1-\frac{k}{n}\right)}} < \pi \\ \implies & \pi - \frac{4}{\sqrt{n}} + O\left(\frac{1}{n^{3/2}}\right) < \frac{1}{n}\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{\frac{k}{n}\left(1-\frac{k}{n}\right)}} < \pi
\end{align*}

Thus, from the upper bound we have, $$\sum\limits_{n \ge 2} \left(\frac{1}{n}\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{\frac{k}{n}\left(1-\frac{k}{n}\right)}}\right)x^{n-2} < \frac{\pi}{(1-x)} \implies \lim\limits_{x \to 1^{-}} \sqrt{1-x}\left(\sum\limits_{n \ge 1} \frac{x^{n-1}}{\sqrt{n}}\right) \le \sqrt{\pi}$$

Again, from the lower bound we have, \begin{align*}& \frac{\pi}{(1-x)} -4 \left(\sum\limits_{n \ge 2} \frac{x^{n-1}}{\sqrt{n}}\right) + O\left(\sum\limits_{n \ge 2} \frac{x^{n-1}}{n^{3/2}}\right) < \sum\limits_{n \ge 2} \left(\frac{1}{n}\sum\limits_{k=1}^{n-1}\frac{1}{\sqrt{\frac{k}{n}\left(1-\frac{k}{n}\right)}}\right)x^{n-2} \\ \implies & \pi - \lim\limits_{x \to 1^{-}} 4(1-x)\sum\limits_{n \ge 2} \frac{x^{n-1}}{\sqrt{n}} + O(1-x) \le \lim\limits_{x \to 1^{-}} (1-x)\left(\sum\limits_{n \ge 1} \frac{x^{n-1}}{\sqrt{n}}\right)^2 \\ \implies & \sqrt{\pi} \le \lim\limits_{x \to 1^{-}} \sqrt{1-x}\left(\sum\limits_{n \ge 1} \frac{x^{n-1}}{\sqrt{n}}\right)
\end{align*}

Thus, $\displaystyle \lim\limits_{x \to 1^{-}} \sqrt{1-x}\left(\sum\limits_{n \ge 1} \frac{x^{n-1}}{\sqrt{n}}\right) = \sqrt{\pi}$

and consequently we have, $$\lim_{x\to-e^+}\int_{0}^{+\infty}\frac{(x+e)^{1/2}}{e^t+xt}\,dt = \frac{1}{\sqrt{e}}\lim\limits_{x \to 1^{-}} \sqrt{1-x}I(x) = \pi\sqrt{\frac{2}{e}} $$