Question on reasoning for $\int_1^\infty\frac{\sin(x)}{x}dx$ to converge

Question

I often saw a 'proof' that $\int_1^\infty\frac{\sin(x)}{x}dx$ converges:

By integration by parts we get $$\int_1^\infty\frac{\sin(x)}{x}dx = \cos(1)-\int_1^\infty{\frac{\cos(x)}{x^2}}dx$$ and thus because $\frac{\cos(x)}{x^2} \leq \frac{1}{x^2}$ and $\int_1^\infty{\frac{1}{x^2}}dx$ converges $$\int_1^\infty{\frac{\cos(x)}{x^2}}dx$$ also converges by comparison test.

But isn't it just wrong to reason like that? Just because $$\lim_{C\to\infty}\int_{1}^C{\frac{1}{x^2}dx}$$ exists and $$\int_{1}^C{\frac{\cos(x)}{x^2}dx} \leq \int_{1}^C{\frac{1}{x^2}dx}$$ for all $C\geq 1$, it dosn't follow that $$\lim_{C\to\infty}\int_{1}^C{\frac{\cos(x)}{x^2}dx}$$ exists. Am I missing something obvious? This 'proof' also appears in the book "Analysis 1" by Königsberger.

Answer 1

Maybe the clearest intuition comes from sequences. Let $(c_n)$ be any increasing sequence which goes to infinity, and consider $$I_n=\int_1^{c_n} \frac{\cos x}{x^2}\,dx.$$ We have $$|I_n-I_m|\le \int_{c_m}^{c_n} \frac{1}{x^2}\,dx,$$ so the sequence $(I_n)$ is Cauchy, and therefore converges.

Answer 2

Here we are not talking about the existence of the integral, we just want to prove whether the integral converges or not. As such, the integral is non integrable. But, if the integral of $\frac{sin(x)}{x}$ exists then the integral of $\frac{cos(x)}{x^2}$ has to exist if the integration by parts approach is considered.

To prove the existence of the integral of $\frac{cos(x)}{x^2}$ you can use Taylor series for $cos(x)$ which gives a finite value of the integral for $\frac{cos(x)}{x^2}$.

As far as convergence is considered comparison test gives $\frac{cos(x)}{x^2}<\frac{1}{x^2}$. The inequality applies over their integrals also. Thus, integral of $\frac{sin(x)}{x}$ is a finite value lying between some finite constraints. Therefore, if integration by parts is true and integral of $\frac{cos(x)}{x^2}$ exists then the integral of $\frac{sin(x)}{x}$ converges. Thus, the proof is correct.

By convergence of the integral I mean that the integral must converge to a finite value.A finite value is a value which lies in finite measurable constraint. For example $\pi$ lies between $3$ and $4$.

I hope that the explanation helps to whatever you are trying understand in the question

Answer 3

The argument in the proof you report is perhaps a bit sloppy. But we can make it more rigorous. For $t\ge1$ we have $$ \int_1^t\frac{\sin x}{x}\,dx= \left[-\frac{\cos x}{x}\right]_1^t- \int_1^t\frac{\cos x}{x^2}\,dx $$ Since we are interested in the limit for $t\to\infty$ and $$ \lim_{t\to\infty}\left[-\frac{\cos x}{x}\right]_1^t = \lim_{t\to\infty}\left(-\frac{\cos t}{t}+\frac{\cos 1}{1}\right) = \cos 1 $$ the limit we are interested in exists and is finite if and only if $$ \lim_{t\to\infty}\int_1^t\frac{\cos x}{x^2}\,dx $$ exists and is finite, that is, if and only if $$ \int_1^\infty\frac{\cos x}{x^2}\,dx $$ converges.

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Now the proof of the convergence.

Consider first two nonnegative continuous functions $f$ and $g$ defined on $[a,\infty)$; if $$ \int_{a}^{\infty}g(x)\,dx $$ converges, then also $$ \int_a^{\infty}f(x)\,dx $$ converges. Indeed, for every $t\ge a$ we have $$ \int_a^t f(x)\,dx\le \int_a^t g(x)\,dx $$ and, since the function $$ G(t)=\int_a^t g(x)\,dx $$ is non decreasing, we have, for all $t\ge a$, $$ \int_a^t g(x)\,dx=G(t)\le\lim_{t\to\infty}G(t)=\int_a^\infty g(x)\,dx $$ Therefore the function $$ F(t)=\int_a^t f(x)\,dx $$ is non decreasing and bounded; hence $$ \lim_{t\to\infty} F(t)=\int_a^\infty f(x)\,dx $$ exists and is finite.

This applies to show that $$ \int_1^\infty\frac{|\cos x|}{x^2}\,dx $$ converges.

Now there's another important theorem. Suppose $f$ is a continuous function on $[a,\infty)$ and that $$ \int_a^{\infty}|f(x)|\,dx $$ converges. Then also $$ \int_a^{\infty}f(x)\,dx $$ converges.

Indeed, consider $f_+(x)=\max\{f(x),0\}$ and $f_-(x)=-\min\{f(x),0\}$. It's easy to see that $f_+$ and $f_-$ are continuous (and nonnegative) and that $$ f(x)=f_+(x)-f_-(x),\quad |f(x)|=f_+(x)+f_-(x) $$ By the previous result, both $$ \int_a^{\infty}f_+(x)\,dx\quad\text{and}\quad\int_a^{\infty}f_-(x)\,dx $$ converge. Now \begin{align} \lim_{t\to\infty}\int_{a}^t f(x)\,dx &= \lim_{t\to\infty}\int_{a}^t (f_+(x)-f_-(x))\,dx\\ &= \lim_{t\to\infty}\left(\int_{a}^t f_+(x)\,dx - \int_{a}^t f_-(x)\,dx\right)\\ &= \lim_{t\to\infty}\int_{a}^t f_+(x)\,dx - \lim_{t\to\infty}\int_{a}^t f_-(x)\,dx \end{align} because both the final limits exist and are finite.