Find the maximum value $\alpha$ for which $\int_{0}^{\frac{1}{\pi}} \frac{1}{x^\alpha} \sin{\frac{1}{x}}dx$ converges.

Question

Many calculus or analysis books write improper integrals $\int_{a}^{b} f(x)dx$, where $f(x)$ is a function such that $\lim_{x\to a} f(x)=+\infty$ or $\lim_{x\to a} f(x)=-\infty$ or $\lim_{x\to b} f(x)=+\infty$ or $\lim_{x\to b} f(x)=-\infty$.

But many calculus or analysis books don't write improper integrals $\int_{a}^{b} f(x)dx$, where $f(x)$ has a dicontinuity of the second kind at $x=a$ or $x=b$.

Why?

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So, I am interested in an improper integral $\int_{a}^{b} f(x)dx$, where $f(x)$ has a dicontinuity of the second kind at $x=a$ or $x=b$.

$\frac{1}{x^\alpha} \sin{\frac{1}{x}}$ has a discontinuity of the second kind at $x=0$.

Find the maximum value $\alpha$ for which $\int_{0}^{\frac{1}{\pi}} \frac{1}{x^\alpha} \sin{\frac{1}{x}}dx$ converges.

$\int_{0}^{\frac{1}{\pi}} \frac{1}{x^2} \sin{\frac{1}{x}}dx$ diverges.
$\int_{0}^{\frac{1}{\pi}} \frac{1}{x^1} \sin{\frac{1}{x}}dx$ converges.
$\int_{0}^{\frac{1}{\pi}} \frac{1}{x^{1.5}} \sin{\frac{1}{x}}dx$ converges.

Answer 1

But many calculus or analysis books don't write improper integrals $\displaystyle\int_{a}^{b} f(x)\,\text{d}x$, where $f(x)$ has a discontinuity of the second kind at $x=a$ or $x=b$.

Why?

I guess the reason has to do with cases where $f(x)$ is not Lebesgue-integrable on $(a,b]$ but where $\displaystyle\lim_{{t\to a}^+}\int_{t}^b f(x)\,\text{d}x$ exists. In such cases the authors of an introductory calculus text might not want to confuse people.

To take an example you cited, $\displaystyle\lim_{{a\to0}^+}\int_{a}^{\frac{1}{\pi}} \frac{1}{x^1} \sin{\frac{1}{x}}dx$ is exactly such a case.

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By the way,

$\displaystyle\int_{a}^{\frac{1}{\pi}} \frac{1}{x^c} \sin{\frac{1}{x}}\,\text{d}x=\int_{\pi}^{1/a} y^{c-2}\sin(y)\,\text{d}y$

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$\displaystyle\lim_{b\to\infty}\int_{\pi}^{b} y^g\sin(y)\,\text{d}y$ exists if $g<0$.

Reference: Prove that $\int_0^{\infty} \frac{\sin x}{x^p}\, dx$ converges for $0<p<2$

However, $\displaystyle\lim_{b\to\infty}\int_{\pi}^{b} y^g\sin(y)\,\text{d}y$ does not exist when $g\geq0$.

Hint: how large is $\displaystyle\int_{n\pi}^{(n+1)\pi}y^g\sin(y)\,\text{d}y$?

Answer 2

I assume that $a$ is real.

$$\int_0^{\frac 1 \pi} {x^{a}} \sin \left(\frac{1}{x}\right)\,dx=\int_\pi^{\infty} t^{a-2} \sin(t)\,dt$$ makes clear that something happens around $a=2$ where the integral does not converge. The same for $a>2$.

Now, for $a<2$, in terms of hypergeometric function $$\int_\pi^{\infty} t^{a-2} \sin(t)\,dt=-\cos \left(\frac{\pi a}{2}\right)\ \Gamma (a-1)-\frac{\pi ^a}{a}\,\, _1F_2\left(\frac{a}{2};\frac{3}{2},\frac{a+2}{2};-\frac{\pi ^2}{4}\right)$$

What is interesting is that, asking Wolfram Alpha to compute the integral for $a=\frac{19}{10}$, the integral is said to not converge why, using the above formula, its value is $-0.884766$ which is confirmed by numerical integration.