Your working is perfectly correct, the only incorrect logic is the following:
I am clearly wrong since Wolfram|Alpha disagrees with me
To compute the integral Wolfram|Alpha has used a numerical method to estimate the value and the method it is using is not sophisticated enough to handle this integral near the asymptote.
This is not uncommon. I can't remember specific instances off the top of my head, but several times I've been able to work out simpler expressions than the forms returned in WA. Humans are still usually better mathematicians than computers (13/02/12) .
$\def\R{\mathbb{R}}\def\N{\mathbb{N}}\def\d{\mathrm{d}}\def\e{\mathrm{e}}\def\peq{\mathrel{\phantom{=}}{}}$The proposition is not necessarily true.
Lemma 1: If $f \in C^โ((0, +โ))$ and $g(x) := f(\e^x)$, then $g \in C^โ(\R)$ and the following conditions are equivalent:
- $\lim\limits_{s โ +โ} s^m f^{(m)}(s) = 0$ for all $m \in \N$;
- $\lim\limits_{x โ +โ} g^{(m)}(x) = 0$ for all $m \in \N$.
(In fact, $g^{(m)}(x)$ is a linear combination of $f(s), s f'(s), \cdots, s^m f^{(m)}(s)$ with $s = \e^x$.)
Lemma 2: If $h \in C^โ((0, +โ))$, then for each $m \in \N_+$, there exist polynomials $P_m, Q_m \in \R[h_1, \cdots, h_m]$ such that\begin{align*}
(\sin(h(x)))^{(m)} &= P_m(h'(x), \cdots, h^{(m)}(x)) \sin(h(x))\\
&\peq + Q_m(h'(x), \cdots, h^{(m)}(x)) \cos(h(x)).\quad \forall x > 0
\end{align*}
(It can be easily proved by induction on $m$.)
Now define $G(x) = \sin(h(x))$, where $h(x) = x^a$ and $a \in (0, 1)$ is a constant, and take $ฮด(s) = G'(\ln s)$. Since $\lim\limits_{x โ +โ} h^{(m)}(x) = 0$ for any $m \in \N_+$, then Lemma 2 yields that $\lim\limits_{x โ +โ} G^{(m)}(x) = 0$ for all $m \in \N_+$, and combining Lemma 1 shows that $\lim\limits_{s โ +โ} s^m ฮด^{(m)}(s) = 0$ for all $m \in \N$.
For $s > r > 1$, making the substitution $u = \e^x$ yields\begin{gather*}
\left| \int_r^s \frac{ฮด(u)}{u} \,\d u \right| = \left| \int_{\ln r}^{\ln s} ฮด(\e^x) \,\d x \right| = \left| \int_{\ln r}^{\ln s} G'(x) \,\d x \right|\\
= |G(\ln s) - G(\ln r)| \leqslant |G(\ln s)| + |G(\ln r)| \leqslant 2.
\end{gather*}
For $B > A > 1$ and any $s > 1$, because\begin{align*}
&\peq |G(\ln(Bs)) - G(\ln(As)))| = |\sin(h(\ln(Bs))) - \sin(h(\ln(Bs)))|\\
&= 2 \left| \sin\left( \frac{1}{2} (h(\ln(Bs)) - h(\ln(As))) \right) \right| ยท \left| \cos\left( \frac{1}{2} (h(\ln(As)) + h(\ln(Bs))) \right) \right|\\
&\leqslant 2 \left| \frac{1}{2} (h(\ln(Bs)) - h(\ln(As))) \right| = |(\ln s + \ln B)^a - (\ln s + \ln A)^a|
\end{align*}
and $\lim\limits_{s โ +โ} ((\ln s + \ln B)^a - (\ln s + \ln A)^a) = 0$, so$$
\lim_{s โ +โ} \int_{As}^{Bs} \frac{ฮด(u)}{u} \,\d u = \lim_{s โ +โ} (G(\ln(Bs)) - G(\ln(As))) = 0.
$$
However, for any $s > s_0 > 1$, since$$
\int_{s_0}^s \frac{ฮด(u)}{u} \,\d u = G(\ln s) - G(\ln s_0)
$$
and $\lim\limits_{x โ +โ} G(x)$ does not exist, then $\displaystyle \int_{s_0}^s \frac{ฮด(u)}{u} \,\d u$ does not exist.
Best Answer
You don't have to care about the exact value of the integral. Simply use asymptotics.
To see how the integrand functions behaves asymptotically as $x \to \infty$, compute the sum of the two fractions.
$$\frac{r}{x+1} - \frac{3x}{2x^2+r} = \frac{(2r-3)x^2 -3x + r^2-1}{(x+1)(2x^2+r)}$$
Now consider two cases:
If $2r-3 = 0$, then asymptotically as $x \to \infty$ $$\frac{(2r-3)x^2 -3x + r^2-1}{(x+1)(2x^2+r)} \sim \frac{-3x}{2x^3} = O (1/x^2)$$ hence the integral is convergent.
If $2r-3 \neq 0$, then asymptotically as $x \to \infty$ $$\frac{(2r-3)x^2 -3x + r^2-1}{(x+1)(2x^2+r)} \sim \frac{(2r-3)x^2}{2x^3} = C \cdot \frac{1}{x}$$ (where $C \neq 0$ is a constant), hence the integral is divergent.
This means that the integral converges only when $r=3/2$.