Limit of an integral (remainder term of a Euler-Maclaurin expansion)

For every $x > 0$, define $$I(x) = \int_1^\infty \left(\{t\} - \frac{1}{2}\right)\frac{x}{e^{xt}-1}\,dt.$$ where $\{x\} = x - \lfloor x\rfloor$ denotes the fractional part of $x$.

How to justify that $I(x)$ converges to the improper integral $$ \int_1^\infty \left(\{t\}-\frac{1}{2}\right)\frac{dt}{t} $$ when $x \searrow 0$ ?

Of course $\dfrac{x}{e^{tx}-1}$ converges to $\dfrac{1}{t}$ for every $t$, but since $$ \int_1^\infty \left|\{t\} - \frac{1}{2}\right|\frac{dt}{t} = \sum_{n=1}^\infty \int_{-1/2}^{1/2} \frac{|u|\,du}{n+\frac{1}{2}+u} \geq \left(\sum_{n=2}^\infty \frac{1}{n}\right)\int_{-1/2}^{1/2}|u|\,du = +\infty, $$ there is no hope to apply Lebesgue's theorem.

I noticed that this is some type of Abelian-like result. Here is a proof.

Let $f(t)=\left(\{t\}-\dfrac{1}{2}\right)\dfrac{1}{t}$. It is known that $F(T) = \int_1^T f(t)\,dt$ converges to $\ell = \dfrac{1}{2}\ln(2\pi)-1$ as $T$ tends to $+\infty$. Using the following integration by parts formula:

$$\int_T^\infty f(t)\frac{tx}{e^{tx}-1}\,dt = -(F(T)-\ell)\frac{Tx}{e^{Tx}-1}-\int_T^\infty (F(t)-\ell)\frac{d}{dt}\frac{tx}{e^{tx}-1}\,dt$$ where $\dfrac{d}{dt}\dfrac{tx}{e^{tx}-1}=\dfrac{xe^{tx}(1-tx-e^{-tx})}{(e^{tx}-1)} \leq 0$, we get $$ \left|\int_T^\infty f(t)\dfrac{tx}{e^{tx}-1}\,dt\right| \leq 2\sup_{t\geq T}|F(t)-\ell|\dfrac{Tx}{e^{Tx}-1}\leq 2 \sup_{t\geq T}|F(t)-\ell|. $$

Hence, for every $T > 1$ and $x >0$, $$ \left|\int_1^\infty f(t)\dfrac{xt}{e^{xt}-1}\,dt - \ell\right| \leq \left|\int_1^T f(t)\left(1-\dfrac{xt}{e^{xt}-1}\right)\,dt\right| + 3\sup_{t\geq T}|F(t)-\ell|. $$ The result easily follows from this inequality.

Solution 1:

Let $$I_0 = \int_{1}^{\infty} \left(\left\{t\right\} - \dfrac12 \right) \dfrac{dt}{t}$$ We then have \begin{align} \vert I(x) - I_0 \vert & = \left \vert \int_{1}^{\infty} \left(\left\{t\right\} - \dfrac12 \right) \left(\dfrac{x}{e^{xt}-1} - \dfrac1t\right) dt\right \vert\\ & = \left \vert \int_{1}^{\infty} \left(\left\{t\right\} - \dfrac12 \right) \left(\dfrac{x}{e^{xt}-1} - \dfrac1t\right) dt\right \vert\\ & = \left \vert \int_{1}^{\infty} \left( \dfrac12 -\left\{t\right\} \right) \left(\dfrac{e^{xt}-1-xt}{t(e^{xt}-1)}\right) dt\right \vert\\ \end{align} Now note that $$\left(\dfrac{e^{xt}-1-xt}{t(e^{xt}-1)}\right) = x\left(\dfrac{e^{xt}-1-xt}{xt(e^{xt}-1)}\right) = x f(xt)$$ where $f(a) = \dfrac{e^a-1-a}{a(e^a-1)}$. Note that $\lim_{a \to 0^+} f(a) =\dfrac12$, $f(a)$ is a decreasing function and $\lim_{a \to +\infty} f(a) = 0$ $$\int_{n}^{n+1/2} \left(\dfrac12 - \{t\}\right)x f(xt) dt = \int_{n}^{n+1/2} \left(\dfrac12 - t + n\right)x f(xt) dt\\ = x f(xt_{n1}) \int_{n}^{n+1/2} \left(\dfrac12 - t + n\right) dt= \dfrac{x f(xt_{n1})}8$$ where $t_{n1} \in [n,n+1/2]$ by MVT. $$\int_{n+1/2}^{n+1} \left(\dfrac12 - \{t\}\right)x f(xt) dt = \int_{n+1/2}^{n+1} \left(\dfrac12 - t + n\right)x f(xt) dt\\ = x f(xt_{n2}) \int_{n+1/2}^{n+1} \left(\dfrac12 - t + n\right) dt = -\dfrac{x f(xt_{n2})}8$$ where $t_{n2} \in [n+1/2,n+1]$ by MVT.

Hence, $$\int_{1}^{\infty} \left( \dfrac12 -\left\{t\right\} \right) \left(\dfrac{e^{xt}-1-xt}{t(e^{xt}-1)}\right) dt = \dfrac{x}8 \left( f(x_{01}) - f(x_{02}) + f(x_{11}) - f(x_{12}) + f(x_{21}) - f(x_{22}) + \cdots \right)$$ Now using the alternating test, since $f(a)$ is a decreasing function converging to $0$,$$\left( f(x_{01}) - f(x_{02}) + f(x_{11}) - f(x_{12}) + f(x_{21}) - f(x_{22}) + \cdots \right)$$ converges to a finite number, say $L$. Hence, we have $$\int_{1}^{\infty} \left( \dfrac12 -\left\{t\right\} \right) \left(\dfrac{e^{xt}-1-xt}{t(e^{xt}-1)}\right) dt = \dfrac{xL}8$$ Now take limit as $x \to 0$ to finish it off.