This is a question I recently got on an exam:
Find whether $\displaystyle \mathcal I=\int_0^1\frac{\sin^3(1/x)\cos(1/x)}{x^{3/2}+x^2\ln(1+1/x)}dx$ converges or diverges.
So I substituted $u=1/x$ to get $$\mathcal I=\int_1^\infty\frac{\sin^3 u\cos u}{\sqrt u+\ln(1+u)}du.$$
Now, take $f(x)=\sin^3x\cos x$ and $g(x)=\dfrac{1}{\sqrt x+\ln(1+x)}$ so that $\bigg|\displaystyle\int_1^b f(x)dx\bigg|$ exists and is uniformly bounded by $2$ $\forall b>1$, and $g(x)$ is non-negative monotone and bounded on $[1,\infty)$ and $\displaystyle \lim_{x\to \infty}g(x)=0$.
Thus, by Dirichlet's test for improper integrals, $\mathcal I$ should converge.
However, when I plugged the integral into Wolfram Alpha, it repeatedly exceeded its standard computation time. Also, when I input it into integral-calculator, it wasn't able to calculate it, and when I selected "integrate numerically only", it showed that "cannot compute because integral may be divergent". However, desmos shows that it is $0$. Thus I am confused. If $\mathcal I$ converges, then shouldn't the softwares be easily able to compute its value?
So is there any other method (comparison test, limit ratio test, $\mu$-test etc.) to confirm the convergence/divergence if this integral?
Best Answer
Your proof of convergence is correct. Note that the Dirichlet's test gives you also upper and lower bounds: by integrating by parts we have that $$I=\int_1^\infty\frac{\sin^3 (u)\cos (u)}{\sqrt u+\ln(1+u)}du=-\frac{\sin^4(1)}{4(1+\ln(2))}+ \frac{1}{4}\int_1^\infty\frac{\sin^4(u)\Big(\frac{1}{2\sqrt{u}}+\frac{1}{1+u}\Big)}{(\sqrt u+\ln(1+u))^2}du.$$ Therefore $$-\frac{\sin^4(1)}{4(1+\ln(2))}<I<-\frac{\sin^4(1)}{4(1+\ln(2))}+\frac{1}{4}\int_1^\infty\frac{\frac{1}{2\sqrt{u}}+\frac{1}{1+u}}{(\sqrt u+\ln(1+u))^2}du=\frac{1-\sin^4(1)}{4(1+\ln(2))}$$ that is $I\in (-0.0740289,0.07362502)$.
A graph of the integral function $x\to \int_1^\infty\frac{\sin^3 (u)\cos (u)}{\sqrt u+\ln(1+u)}du$, via Desmos, shows a slow oscillating curve.
For a more precise numerical approximation, compute $$I_R=-\frac{\sin^4(1)}{4(1+\ln(2))}+\frac{1}{4}\int_1^R\frac{\sin^4(u)\Big(\frac{1}{2\sqrt{u}}+\frac{1}{1+u}\Big)}{(\sqrt u+\ln(1+u))^2}du$$ with $R=N\pi$ with $N$ positive integer, then the error can be estimated as $$0<I-I_R=\frac{1}{4}\int_{R}^{\infty}\sin^4(u)f(u)du= \frac{1}{4}\sum_{k=N}^{\infty}\int_{0}^{\pi}\sin^4(k\pi +u)f(k\pi +u)du\\ \leq \frac{1}{4}\sum_{k=N}^{\infty}f(k\pi)\int_{0}^{\pi}\sin^4(u) du =\frac{3\pi}{32}\sum_{k=N}^{\infty}f(k\pi) $$ where $f(u)=\frac{\frac{1}{2\sqrt{u}}+\frac{1}{1+u}}{(\sqrt u+\ln(1+u))^2}$ is decreasing and positive.
Using Maple, for $R=10000\pi$, I got that $I_R\approx -0.001744631$ with $I-I_R<0.000499744$, that is $I\in(-0.001744631,-0.001244887)$.