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Prove that $$ \int_{0}^{\infty} \frac{\sin^2 x-x\sin x}{x^3} \, dx = \frac{1}{2} - \ln 2 .$$

Integration by parts gives

\begin{align*} &\lim_{R\to \infty} \int_{0}^{R} \frac{\sin^2 x-x\sin x}{x^3} \, dx \\ &= \lim_{R\to \infty} \biggl( \int_{0}^{R} \frac{\sin^2x}{x^3} \, dx - \int_{0}^{R} \frac{\sin x}{x^2} \, dx \biggr)\\ &= \lim_{R\to\infty} \biggl( \frac{\sin^2 x}{-2x^2}\Biggr\rvert_{0}^{R} - \int_{0}^{R} \frac{\sin (2x)}{-2x^2} \, dx - \biggl(-\frac{\sin x}{x} \Biggr\rvert_{0}^{R} + \int_{0}^{R} \frac{\cos x}{x} \,dx \biggr) \biggr) \\ &= \lim_{R\to \infty} \biggl(\frac{1}{2} + \int_{0}^{2R} \frac{\sin u}{u^2/2} \, \Bigl(\frac{1}2 \, du\Bigr) - \biggl( 1 + \int_{0}^{R} \frac{\cos x}{x} \, dx \biggr)\biggr) \\ &\hspace{22em}\text{(using the substitution $u\mapsto 2x$)}\\ &= -\frac{1}{2} + \lim_{R\to \infty} \biggl(\int_{0}^{2R} \frac{\sin u}{u^2} \, du - \int_{0}^{R} \frac{\cos x}{x} \,dx \biggr)\\ &= \frac{1}{2} + \lim_{R\to\infty} \biggl(\int_{R}^{2R} \frac{\cos x}{x} \, dx \biggr) \end{align*}

Thus it suffices to show that $\lim_{R\to\infty} \int_{R}^{2R} \frac{\cos x}{x} \, dx = \ln 2$. The Taylor series expansion of $\cos x$ is given by $\cos x = \sum_{i=0}^{\infty} \frac{(-1)^i x^{2i}}{(2i)!}$.

(If the step below (the one involving the interchanging of an infinite sum and integral) is valid, why exactly is it valid? For instance, does it use uniform convergence?)

The limit equals

$$ \lim_{R\to\infty} \int_{R}^{2R} \biggl( \frac{1}{x} + \sum_{i=1}^{\infty} \frac{(-1)^i x^{2i-1}}{(2i)!} \biggr) \, dx = \ln 2 + \lim_{R\to\infty} \sum_{i=1}^{\infty} \biggl[\frac{(-1)^ix^{2i}}{(2i)(2i)!}\biggr]_{R}^{2R} . $$

But I don't know how to show $\lim_{R\to\infty} \lim_{R\to\infty} \sum_{i=1}^{\infty} \Bigl[\frac{(-1)^ix^{2i}}{(2i)(2i)!}\Bigr]_{R}^{2R} = -2 \ln 2$.

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    $\begingroup$ Taylor series is notoriously hard to use for studying global behavior of the function it defines. Also, note that you cannot split the original integral into two parts, as both separated integrals diverge. In fact, you can check that $$\lim_{R\to\infty}\int_{R}^{2R}\frac{\cos x}{x}\,\mathrm{d}x=0$$ is not the correct source of that logarithm term, and the correct source can only be identified via a proper treatment of the singularities of the integrands about the origin. $\endgroup$ Nov 14, 2021 at 3:05
  • $\begingroup$ Writing $\int_0^R {\frac{{\sin ^2 x}}{{x^3 }}dx}$ and $\int_0^R {\frac{{\sin x}}{{x^2 }}dx}$ is meaningless. The integrals are divergent, because of the non-integrable singularities of the integrands at $x=0$. $\endgroup$
    – Gary
    Nov 14, 2021 at 3:27

3 Answers 3

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Here is a correct solution that you might want to compare with:

\begin{align*} &\int_{0}^{\infty}\frac{\sin^2 x - x \sin x}{x^3} \, \mathrm{d}x \\ &= \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \int_{\varepsilon}^{R}\frac{\sin^2 x - x \sin x}{x^3} \, \mathrm{d}x \\ &= \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \biggl( \left[ -\frac{\sin^2 x}{2x^2} + \frac{\sin x}{x} \right]_{\varepsilon}^{R} + \int_{\varepsilon}^{R} \left( \frac{\sin(2x)}{2x^2} - \frac{\cos x}{x} \right) \, \mathrm{d}x \biggr) \\ &= -\frac{1}{2} + \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \biggl( \int_{2\varepsilon}^{2R} \frac{\sin x}{x^2} \, \mathrm{d}x - \int_{\varepsilon}^{R} \frac{\cos x}{x} \, \mathrm{d}x \biggr) \\ &= -\frac{1}{2} + \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \biggl( \left[ -\frac{\sin x}{x} \right]_{2\varepsilon}^{2R} + \int_{2\varepsilon}^{2R} \frac{\cos x}{x} \, \mathrm{d}x - \int_{\varepsilon}^{R} \frac{\cos x}{x} \, \mathrm{d}x \biggr) \\ &= \frac{1}{2} + \lim_{\substack{R\to\infty \\ \varepsilon \to 0^+}} \biggl( \int_{R}^{2R} \frac{\cos x}{x} \, \mathrm{d}x - \int_{\varepsilon}^{2\varepsilon} \frac{\cos x}{x} \, \mathrm{d}x \biggr). \end{align*}

Now by noting that

\begin{align*} \left|\int_{R}^{2R} \frac{\cos x}{x} \, \mathrm{d}x\right| &= \left| \frac{\sin R}{R} - \frac{\sin (2R)}{2R} + \int_{R}^{2R} \frac{\sin x}{x^2} \, \mathrm{d}x \right| \\ &\leq \frac{1}{R} + \frac{1}{2R} + \int_{R}^{2R} \frac{1}{x^2} \, \mathrm{d}x \\ &= \frac{2}{R}, \end{align*}

it follows that

$$ \lim_{R\to\infty} \int_{R}^{2R} \frac{\cos x}{x} \, \mathrm{d}x = 0. $$

On the other hand, by substituting $x = \varepsilon u$,

$$ \int_{\varepsilon}^{2\varepsilon} \frac{\cos x}{x} \, \mathrm{d}x = \int_{1}^{2} \frac{\cos(\varepsilon u)}{u} \, \mathrm{d}u \xrightarrow{\varepsilon \to 0^+} \int_{1}^{2} \frac{1}{u} \, \mathrm{d}u = \log 2 $$

by the dominated convergence theorem. Therefore

$$ \int_{0}^{\infty}\frac{\sin^2 x - x \sin x}{x^3} \, \mathrm{d}x = \frac{1}{2} - \log 2. $$

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$$I=\int \frac{\sin^2 (x)-x\sin (x)}{x^3}\, dx=\int \frac{\sin^2 (x)}{x^3}\, dx-\int \frac{\sin (x)}{x^2}\, dx$$ $$\int \frac{\sin^2 (x)}{x^3}\, dx=\frac 1 2\int \frac{1-\cos(2x)}{x^3}\,dx$$ $$\int \frac{1-\cos(2x)}{x^3}\,dx=-\frac 1{2x^2}-\int \frac{\cos(2x)}{x^3}\,dx$$ $$\int \frac{\cos(2x)}{x^3}\,dx=-\frac{\cos(2x)}{x^2}-\int\frac{\sin(2x)}{x^2}\,dx$$ $$\int\frac{\sin(2x)}{x^2}\,dx=-\frac{\sin(2x)}{x}+2\int \frac{\cos(2x)}x\,dx$$ $$\int \frac{\cos(2x)}x\,dx=\int \frac{\cos(y)}y\,dy=\text{Ci}(y)=\text{Ci}(2x)$$

Combining all the above $$\color{red}{\int \frac{\sin^2 (x)}{x^3}\, dx=\text{Ci}(2 x)-\frac{\sin (x) (\sin (x)+2 x \cos (x))}{2 x^2}}$$ $$\int \frac{\sin (x)}{x^2}\, dx=-\frac{\sin (x)}{x}+\int \frac{\cos(x)}x\,dx$$

$$\color{red}{\int \frac{\sin (x)}{x^2}\, dx=-\frac{\sin (x)}{x}+\text{Ci}(x)}$$

So, $$\color{blue}{I=-\text{Ci}(x)+\text{Ci}(2 x)-\frac{1}{4 x^2}+\frac{\cos (2 x)}{4 x^2}+\frac{\sin(x)}{x}-\frac{\sin (2 x)}{2 x}}$$ If $x\to \infty$ its value is $0$.

Now, for $x\to 0$, using expansion we have $$\left(\log (2)-\frac{1}{2}\right)-\frac{x^2}{12}+\frac{13 x^4}{1440}+O\left(x^6\right)$$ $$J=\int_\epsilon^\infty \frac{\sin^2 (x)-x\sin (x)}{x^3}\, dx=\left(\frac{1}{2}-\log (2)\right)+\frac{\epsilon^2}{12}+O\left(\epsilon^4\right)$$

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Alternatively, we could use the fact that $$\int_{0}^{\infty} t^{2} e^{-xt} \, \, \mathrm dt = \frac{2}{x^{3}}, \quad x >0,$$ and then apply Fubini's theorem.$$\begin{align} \int_{0}^{\infty} \frac{\sin^{2}x - x \sin x}{x^{3}} \, \mathrm dx &= \frac{1}{2}\int_{0}^{\infty} \left(\sin^{2} x - x \sin x \right) \int_{0}^{\infty} t^{2}e^{-xt} \, \mathrm dt \, \mathrm dx \\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \int_{0}^{\infty} \left(\sin^{2} x - x \sin x \right) e^{-xt} \, \mathrm dx \, \mathrm dt \\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \left(\frac{1}{2} - \frac{\cos 2x}{2} - x \sin x \right) e^{-xt} \, \mathrm dx \, \mathrm dt \\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \left(\frac{1}{2t} - \frac{1}{2} \frac{t}{t^{2}+4}+ \frac{\mathrm d}{\mathrm d t}\frac{1}{t^{2}+1} \right) \, \mathrm dt \tag{1}\\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \left(\frac{1}{2t} - \frac{1}{2} \frac{t}{t^{2}+4}- \frac{2t}{(t^{2}+1)^{2}} \right) \, \mathrm dt \\ &= \frac{1}{2} \int_{0}^{\infty} t^{2}\left(\frac{2}{t(t^{2}+4)} - \frac{2t}{(t^{2}+1)^{2}} \right) \, \mathrm dt \\ &= \frac{1}{2} \int_{0}^{\infty} \left(\frac{2t}{t^{2}+4} - \frac{2t}{t^{2}+1} + \frac{2t}{(t^{2}+1)^{2}} \right) \, \mathrm dt \\ &= \frac{1}{2} \left(\ln \left(\frac{t^{2}+4}{t^{2}+1} \right) - \frac{1}{t^{2}+1} \right)\Bigg|_{0}^{\infty} \\ &= \frac{1}{2} \left(-\ln 4 +1 \right) \\ &= \frac{1}{2} - \ln 2. \end{align} $$


$(1)$ $\int_{0}^{\infty} x \sin(x) e^{-xt} \, \mathrm dx = -\frac{\mathrm d}{\mathrm dt} \int_{0}^{\infty} \sin(x) e^{-xt} \, \mathrm dx$

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  • $\begingroup$ Thanks. But could you explain how you got from the line before (1) in your main proof to (1)? I.e. the part that reads $\frac{1}2\int_0^\infty t^2(1/2 - \cos(2x)/2 ...)dxdt$ to (1). $\endgroup$
    – user3472
    Nov 19, 2021 at 3:00
  • $\begingroup$ @user3472 $$ \begin{align} & \frac{1}{2}\int_{0}^{\infty} t^{2}\int_{0}^{\infty} \left (\frac{1}{2} - \frac{\cos 2x}{2} - x \sin x \right) e^{-xt} \, \mathrm dx \, \mathrm dt \\ &= \frac{1}{2}\int_{0}^{\infty} t^{2} \left[\frac{1}{2} \int_{0}^{\infty}e^{-xt} \, \mathrm dx - \frac{1}{2} \int_{0}^{\infty} \cos(2x) e^{-xt} \, \mathrm dx - \int_{0}^{\infty} x \sin(x) e^{-xt} \, \mathrm dx\right] \, \mathrm dt \end{align}$$ $\endgroup$ Nov 19, 2021 at 3:30
  • $\begingroup$ Thanks. I get it now. $\endgroup$
    – user3472
    Nov 19, 2021 at 3:39

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