Check the convergence of the improper integrals $\int_2^{\infty} \frac{\sin x}{\log x} d x$ and $\int_0^{\infty}\left(\sin x^2\right) d x$

Question

Actually, I was solving a multiple select question:

Which of the following improper integral are convergent?

(1) $\int_0^{\infty} \frac{\sin x}{x} d x$

(2) $\int_2^{\infty} \frac{\sin x}{\log x} d x$

(3) $\int_0^{\infty}\left(\sin x^2\right) d x$

(4) $\int_0^{\infty} e^{-x^2} d x$

For 1st option, I have seen the convergence of $\int_0^\infty \frac{\sin x}{x}dx$ using Laplace transforms to get the integral as $\frac{\pi}{2}$,

For 2 nd option, I got $\int_2^{\infty} \frac{\sin x}{\log x} d x$ is not absolutely convergent since $\frac{|\sin x|}{\log x} \geq \frac{|\sin x|}{x}$ for $x \geq 2$ and $\int_2^{\infty} \frac{|\sin x|}{x} d x$ diverges, but I couldn't conclude about the convergence,

Despite of the derivation of Fresnel function, How to deal with options 3 using the traditional convergence checking methods like comparison, limit comparison, etc?

For 4 th option, I have a fair inequality $e^{-x^2} < e^{-x}$ for $x>1.$

Thanks in advance for the help.

Answer 1

The second integral can be shown to converge in a variety of ways. In particular, it can be shown to converge by some very general approaches that have broad applicability. (This is also applicable to the first and third integrals).

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Dirichlet test for convergence of an improper integral (under weak conditions)

Suppose $f:[a,\infty) \to \mathbb{R}$ is Riemann integrable on any bounded interval such that $\int_a^b f(x) \, dx $ is bounded for all $b > a$ and that the function $g:[a,\infty) \to \mathbb{R}$ is monotonically decreasing to zero as $x \to \infty$. Then the improper integral $\int_a^\infty f(x) g(x) \, dx$ is convergent.

In this case $f(x) = \sin x$ and $g(x) = [\log x]^{-1}$ are easily seen to satisfy the conditions for the Dirichlet test.

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Cauchy criterion and second integral mean value theorem.

(These are more fundamental results that can be used to prove the Dirichlet test.)

The Cauchy criterion states that the improper integral $\int_a^b h(x) \, dx$ is convergent if and only if for any $\epsilon > 0$ there exists $K \geqslant a$ such that $\left| \int_\alpha ^\beta h(x) \, dx\right| < \epsilon$ for all $\beta > \alpha > K$.

A special case of the second integral mean value theorem for integrals is if $f:[a,b) \to \mathbb{R}$ is Riemann integrable and $g:[a,b] \to \mathbb{R}$ is non-negative and decreasing, then there exists $\xi \in [a,b]$ such that $\int_a^b f(x) g(x) \, dx = g(a)\int_a^\xi f(x) \, dx$.

For the problem at hand, we have by the second integral mean value theorem with $f(x) = \sin x$ and $g(x) = [\log x]^{-1}$,

$$\left|\int_\alpha^\beta \frac{\sin x}{\log x} \, dx \right| = \left|\int_\alpha^\beta \frac{\sin x}{\log x} \, dx \right| =\frac{1}{|\log \alpha|}\left|\int_\alpha^\xi \sin x \, dx \right|= \frac{|\cos \alpha - \cos \xi|}{|\log \alpha|}\leqslant \frac{2}{|\log \alpha|}$$

For any $\epsilon >0$ there exists $K$ such that if $\alpha > K$ then $|\log \alpha | = \log \alpha > \frac{2}{\epsilon}$. Hence, with $\beta > \alpha > K$ we have

$$\left|\int_\alpha^\beta \frac{\sin x}{\log x} \, dx \right| < \epsilon, $$

and the improper integral converges by the Cauchy criterion.

Answer 2

$$\int_0^{\infty} \dfrac{\sin x}{x} \mathrm{d} x=\int_0^{1}\dfrac{\sin x}{x} \mathrm{d}x +\int_1^{\infty} \dfrac{\sin x}{x} \mathrm{d} x\\ \int_1^{\infty} \dfrac{\sin x}{x} \mathrm{d} x=-\dfrac{\cos x}{x}\Bigg|^{+\infty}_1-\int_1^{\infty} \dfrac{\cos x}{x^2} \mathrm{d} x$$ Since $|\cos x|\le 1$, $\,\,\displaystyle{\int_0^{\infty} \dfrac{\cos x}{x^2} \mathrm{d} x}\,\,$ converges by comparison test.

$$\int_1^{\infty}\left(\sin x^2\right) \mathrm{d} x=\dfrac{1}{2}\int_1^{\infty} \dfrac{\sin t}{\sqrt{t}} \mathrm{d} t$$ Then use integration by parts as in the first integral.

$$\int_0^{\infty} e^{-x^2} \mathrm{d} x$$ Since $e^t\geq 1+t \quad \forall t \in [0,+\infty) \implies \dfrac{1}{1+x^2}\geq \dfrac{1}{e^{x^2}}$. It follows $$\int_{0}^\infty e^{-x^2}\mathrm{d}x \leq \int_{0}^\infty \dfrac{1}{1+x^2}=\dfrac{\pi}{2}$$

Answer 3

Since you already solved (1) and (4), let us concentrate on the two integrals (2) and (3) of your title.

The convergence of the Fresnel integral (3) is equivalent (by change of variable $y=x^2$) to the convergence of $\int_0^\infty\frac{\sin y}{\sqrt y}\,\mathrm dy.$

Let us now prove that for any decreasing $C^1$ function $f$ vanishing at infinity, $\int_2^{\infty}f(x)\sin x\,\mathrm dx$ converges. This (applied to $f(x)=\frac1{\ln x}$ or to $f(x)=\frac1{\sqrt x}$) will prove the convergence of the two integrals (2) and (3). The recipe is: integration by parts followed by absolute convergence.

$$\begin{align}\int_2^\infty f(x)\sin x\,\mathrm dx&=\left[-f(x)\cos x\right]_2^\infty+\int_2^\infty f'(x)\cos x\,\mathrm dx\\ &=f(2)\cos2+\int_2^\infty f'(x)\cos x\,\mathrm dx \end{align}$$ and $$\begin{align}\int_2^\infty|f'(x)\cos x|\,\mathrm dx&\le\int_2^\infty|f'(x)|\,\mathrm dx\\ &=\int_2^\infty-f'(x)\,\mathrm dx\\ &=\left[-f(x)\right]_2^\infty\\ &=f(2)<\infty. \end{align}$$