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Suppose $ \alpha, \beta>0 $. Compute: $$ \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx $$

Here is what I do: $$\begin{align} \int_{0}^{\infty}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx &= \int_{0}^{\infty}dx\int_{\alpha}^{\beta}\sin (yx)dy\\ &=\int_{\alpha}^{\beta}dy\int_{0}^{\infty}\sin(yx)dx\\ & \\ & \qquad\text{let $ yx=u $}\\ & \\ &=\int_{\alpha}^{\beta}\frac{1}{y}dy\int_0^{\infty}\sin u du\\ &=\int_{\alpha}^{\beta}\frac{1}{y}dy\left( -\cos u|_{\infty}+\cos u|_0 \right)\\ &=\log\frac{\beta}{\alpha}(-\cos(\infty)+1)\\ &=\log\frac{\beta}{\alpha}-\cos(\infty)\log\frac{\beta}{\alpha} \end{align}$$

But $ \cos(\infty) $ does not exist right? Does it mean the integral actually diverse?

Edit: The question comes from https://math.uchicago.edu/~min/GRE/files/week1.pdf

Who can point out my mistake in the above deduction?

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    $\begingroup$ Use Laplace transform or add an artificial term $e^{-tx}$ in your integral. $\endgroup$ – Nosrati Oct 22 '18 at 13:40
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    $\begingroup$ Hi \begin{align} \int_{0}^{\infty}e^{-tx}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx &= \int_{0}^{\infty}dx\int_{\alpha}^{\beta}e^{-tx}\sin (yx)dy\\ &=\int_{\alpha}^{\beta}dy\int_{0}^{\infty}e^{-tx}\sin(yx)dx\\ \end{align} $\endgroup$ – Nosrati Oct 22 '18 at 13:46
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\begin{align} \int_{0}^{\infty}e^{-tx}\frac{\cos (\alpha x)-\cos (\beta x)}{x}dx &= \int_{0}^{\infty}dx\int_{\alpha}^{\beta}e^{-tx}\sin (yx)dy\\ &=\int_{\alpha}^{\beta}dy\int_{0}^{\infty}e^{-tx}\sin(yx)dx\\ &=\int_{\alpha}^{\beta}dy\dfrac{y}{t^2+y^2}\\ &=\dfrac12\ln\dfrac{t^2+\beta^2}{t^2+\alpha^2} \end{align} now let $t=0$.

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    $\begingroup$ You need to justify your interchanging the order of integration. $\endgroup$ – Mark Viola Oct 22 '18 at 14:50
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This is a Frullani integral. Compute as follows.

Let $0 < r < R < +\infty$. Then $$\newcommand \diff {\,\mathrm d} \int_r^R \frac {\cos(\alpha x) - \cos(\beta x)}x \diff x = \int_r^R \frac {\cos(\alpha x)}x \diff x - \int_r^R \frac {\cos(\beta x)}x \diff x =\left( \int_{\alpha r}^{\alpha R} - \int_{\beta r}^{\beta R}\right)\frac {\cos t}t \diff t = \int_{\alpha r}^{\beta r} \frac {\cos t}t\diff t - \int_{\alpha R}^{\beta R} \frac {\cos t} t \diff t = I(r) - J(R). $$ Now for $I(r)$, use the 1st MVT for integrals, we have $$ I(r) = \cos(A) \int_{\alpha r}^{\beta r} \frac {\diff t} t = \cos(A) \log(\beta/\alpha) [A = \alpha r + (1-s)(\beta - \alpha)r, s \in (0,1)] \xrightarrow{r \to 0^+} \cos 0 \log(\beta /\alpha) = \log(\beta/\alpha). $$ For $J(R)$, note that the integral $$ \int_1^{+\infty}\frac {\cos t}t \diff t $$ converges by Dirichlet test, hence $$ J(R) \xrightarrow{R \to +\infty} 0 $$ by Cauchy principle. Altogether the original integral is $$ \lim_{\substack {r \to 0^+\\ R\to +\infty }} I(r) - J(R) = \log\left( \frac \beta \alpha\right). $$

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  • $\begingroup$ To whoever edited my post, thanks for altering the name of the person! I am not a native, so the textbook I used might give the wrong spelling. Anyway, thanks! $\endgroup$ – xbh Oct 23 '18 at 2:55
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I would actually use Laplace transforms to compute this sort of an integral. You must have used it in solving linear differential equations in the past. It can be defined as follows:- $$ \mathcal{L}\{f(x)\}=\int_{0}^\infty e^{-px}f(x)dx = F(p)$$ Here we are transforming the function $f$ with domain $x$ to a function $F$ with domain $p$ with a unilateral integral transform of kernel $e^{-px}$. Now consider the general Laplace transform formula given above. Differentiating both sides with respect to $p$ we get: $$F'(p)=\int_{0}^\infty e^{-px}(-x)f(x)dx=-\mathcal{L}\{xf(x)\} \rightarrow (1)$$ Now put $G(p)$ as Laplace transform of $f(x)/x$ and obtain its differentiation using equation $(1)$:- $$G(p)=\mathcal{L}\left\{\frac{f(x)}{x}\right\}\Rightarrow G'(p)=-\mathcal{L}\{f(x)\}=-F(p)\rightarrow (2)$$ Using the Fundamental Theorem of Calculus (relationship between derivative and integral) for $(2)$: $$G(p)=-\int_{a}^p F(p)dp \Rightarrow \int_{0}^\infty e^{-px}\frac{f(x)}{x}dx=-\int_{a}^p F(p)dp \rightarrow (3)$$ Note that $a$ here is some constant. If $G(p) \rightarrow 0$ as $p \rightarrow \infty$ then we put $a = \infty$ and obtain the following:- $$\int_{0}^\infty e^{-px}\frac{f(x)}{x}dx=\int_{p}^\infty F(p)dp \rightarrow (4)$$ If we let $p \rightarrow 0$ on both sides of equation $(4)$ we get the following: $$\int_{0}^\infty \frac{f(x)}{x}dx=\int_{0}^\infty F(p)dp \rightarrow (5)$$ This is useful for us for finding the improper integral of various functions of the form $f(x)/x$ where the transform $F(p)$ is known. Now I leave the proof for the following upto you (which is elementary considering we use integration by parts): $$\mathcal{L}\{ \cos bx \} = \int_{0}^\infty e^{-px}(\cos bx) dx = \frac{p}{p^2 + b^2} \ (p>0) \rightarrow (6)$$ For some constant $b$. Now using equation (5) and (6) we get: $$\int_{0}^\infty \frac{\cos bx}{x}dx=\int_{0}^\infty \frac{p}{p^2 + b^2}dp \rightarrow (7)$$ Now plugging in equation $(7)$ into the integral we are required to compute:- $$I=\int_{0}^\infty \frac{\cos \alpha x - \cos \beta x}{x}dx = \int_{0}^\infty p \left( \frac{1}{p^2 + \alpha^2} - \frac{1}{p^2 + \beta^2} \right)dp$$ $$\Rightarrow I=\frac{\beta^2-\alpha^2}{2} \int_{0}^\infty \frac{2p}{(p^2+\alpha^2)(p^2 +\beta^2)}dp \rightarrow (8)$$ Set $v=\frac{\beta^2+\alpha^2}{2}; u=\frac{\beta^2-\alpha^2}{2}; t=p^2+u$ and using the substitutions and some further simplification: $$I=\int_{v}^\infty \frac{u}{t^2-u^2}dt = \left[\frac{1}{2}\ln \left|\frac{t-u}{t+u}\right|\right]_{t=v}^{t=\infty}=\frac{1}{2}\ln \left|\frac{u+v}{u-v}\right|$$ Substituting the variables back and rewriting the main equation for $I$ we get: $$\int_{0}^\infty \frac{\cos \alpha x - \cos \beta x}{x}dx = \ln \frac{\beta}{\alpha}$$

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