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$$I=\large \int_{0}^{\infty}\frac{\cos(x/a)-1}{x}\cdot\cos(x)\ln(x)\mathrm dx$$

Attempt:

Split them into

$$\large \int_{0}^{\infty}\frac{\cos(x/a)}{x}\cdot\cos(x)\ln(x)\mathrm dx-\int_{0}^{\infty}\frac{1}{x}\cdot\cos(x)\ln(x)\mathrm dx=J-K$$

Applying integration by parts, $\large u=\cos(x)$

$\large \mathrm du=-\sin(x)$

$\large \mathrm dv=\frac{\ln(x)}{x}\mathrm dx$

$\large v=\frac{\ln^2(x)}{2}$

$$K=\frac{\cos(x)\ln^2(x)}{2}+\frac{1}{2}\int\sin(x)\ln^2(x)\mathrm dx$$

Try another integration by parts,

$\large u=\sin(x)\ln(x)$

$\large \mathrm du=\frac{}\sin(x){x}+\cos(x)\ln(x)$

$\large \mathrm dv=\sin(x)\mathrm dx$

$\large v=\sin(x)$

$$\int\sin(x)\ln^2(x)\mathrm dx=-\cos(x)\sin(x)\ln(x)+\int \left(\frac{\sin(2x)}{2x}+\ln(x)-\sin^2(x)\ln(x)\right)\mathrm dx$$

$$\int\sin(x)\ln^2(x)\mathrm dx=-\cos(x)\sin(x)\ln(x)+x\ln(x)-x+\int \left(\frac{\sin(2x)}{2x}-\sin^2(x)\ln(x)\right)\mathrm dx$$

I have try doing integration by parts to reduce it but, it is not working.

I have try and look up standard table of integrals but can't find much to help me.

I am unable to proceed. Can anyone please lead the way?

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  • $\begingroup$ I have a problem with $K$ which can not integrated if the lower bound is $0$ $\endgroup$ – Claude Leibovici Jul 20 '18 at 8:47
  • $\begingroup$ sorry for the mistake, the limit is $\int_{0}^{\infty}$ $\endgroup$ – user565198 Jul 20 '18 at 8:53
  • $\begingroup$ The problem remains with $K$ around $x=0$. I do not see the trick but I suppose that we must not split as $I=J-K$. $\endgroup$ – Claude Leibovici Jul 20 '18 at 9:00
  • $\begingroup$ When I read answers such as the ones you received, I realize how bad I am ! $\endgroup$ – Claude Leibovici Jul 21 '18 at 2:07
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Lemma. For $a > 0$, we have $$ f(a) := \int_{0}^{\infty} \frac{\cos (ax) - \cos x}{x}\log x \, dx = \frac{(2\gamma + \log a)\log a}{2}. $$

Proof. Consider the function $I(s) = \int_{0}^{\infty} \frac{\cos(ax) - \cos x}{x^{1+s}} \, dx$. Then $I(s)$ defines an analytic function on the strip $\operatorname{Re}(s) \in (-1, 1)$. Also, for $s \in (-1, 0)$ we have

\begin{align*} I(s) &= \int_{0}^{\infty} \left( \frac{1}{\Gamma(1+s)} \int_{0}^{\infty} u^s e^{-xu} \, du \right) (\cos(ax) - \cos x) \, dx \\ &= \frac{1}{\Gamma(1+s)} \int_{0}^{\infty} u^s \left( \int_{0}^{\infty} e^{-xu} (\cos(ax) - \cos x) \, dx \right) \, du \\ &= \frac{1}{\Gamma(1+s)} \int_{0}^{\infty} \left( \frac{u^{s+1}}{u^2+a^2} - \frac{u^{s+1}}{u^2 + 1} \right) \, du \\ &= \frac{(a^s - 1)\mathrm{B}(\frac{s}{2}+1,-\frac{s}{2})}{2\Gamma(1+s)} \\ &= -\frac{\pi (a^s - 1)}{2\sin(\frac{\pi s}{2})\Gamma(1+s)}. \end{align*}

This formula continues to hold on all of the strip by the principle of analytic continuation. So it follows that

$$ f(a) = -I'(0) = -\lim_{s \to 0} I'(s) = \frac{(2\gamma + \log a)\log a}{2}. $$


Using the lemma and $\cos(x/a)\cos x = \frac{1}{2}\left( \cos (\frac{a+1}{a}x) + \cos(|\frac{a-1}{a}|x) \right)$, we obtain

$$ \int_{0}^{\infty} \frac{\cos (ax) - 1}{x}\cos x\log x \, dx = \frac{f\left(\frac{a+1}{a}\right) + f\left(\left|\frac{a-1}{a}\right|\right)}{2}. $$

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Let $b=\frac{1}{a} > 0$ to avoid fractions. Using trigonometric identities we have $$ [\cos(b x) - 1]\cos(x) = - \sin\left(\frac{b}{2}x\right) \left\{\sin\left[\left(\frac{b}{2}-1\right)x\right] + \sin\left[\left(\frac{b}{2}+1\right)x\right]\right\} \, . $$ In my answer to this question I have shown that $$ \int \limits_0^\infty \frac{\sin(\alpha x) \sin(\beta x)}{x} \ln(x) \, \mathrm{d} x = \frac{1}{4} \ln \left(\frac{\lvert \alpha - \beta\rvert}{\alpha + \beta}\right)[\ln(\lvert \alpha^2-\beta^2\rvert) + 2 \gamma]$$ holds for $\alpha, \beta > 0$ if $\alpha \neq \beta$ . Therefore we obtain for $a \neq 1 \Leftrightarrow b \neq 1$ \begin{align} I &= \int \limits_0^\infty \frac{[\cos(b x)-1]\cos(x)}{x} \ln(x) \, \mathrm{d} x \\ &= - \int \limits_0^\infty \frac{\sin\left(\frac{b}{2}x\right) \left\{\sin\left[\left(\frac{b}{2}-1\right)x\right] + \sin\left[\left(\frac{b}{2}+1\right)x\right]\right\}}{x} \ln(x) \, \mathrm{d} x \\ &= \frac{1}{4} \ln(\lvert 1-b\rvert) [\ln(\lvert 1-b\rvert) + 2 \gamma] + \frac{1}{4} \ln( 1+b) [\ln( 1+b) + 2 \gamma] \\ &= \frac{1}{4} [\ln^2 (\lvert 1-b\rvert) + \ln^2 (1+b) + 2 \gamma \ln (\lvert 1-b^2 \rvert)] \, . \end{align}

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This is not an answer but it is too long for a comment.

As I wrote in a comment, I suppose that there is a trick but I suppose that we must not split as $I=J−K$ since there is a serious problem with $K$ around $x=0$.

Close to zero $$\frac{\cos \left(\frac{x}{a}\right)-1}{x}\,\cos (x)\,\log (x)=-\frac{x \log (x)}{2 a^2}+O\left(x^3\right) $$ $$\int \frac{\cos \left(\frac{x}{a}\right)-1}{x}\,\cos (x)\,\log (x)\,dx=\frac{x^2 (1-2 \log (x))}{8 a^2}+O\left(x^4\right)$$ does not make any problem.

Being short of ideas and lazy, just by curiosity, I gave it to a CAS and obtained the result $$I=\frac{1}{16} \left(\log ^2\left(\frac{(a-1)^2}{a^2}\right)+4 \gamma \log \left(\frac{(a-1)^2}{a^2}\right)+4 \log \left(\frac{1+a}{a}\right) \left(\log \left(\frac{1+a}{a}\right)+2 \gamma \right)\right)$$ How to get it, this is the question !

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  • $\begingroup$ I have simplified to $\frac{\ln^2(\frac{a}{a-1})-\ln^2(\frac{a}{a+1})+2\gamma\ln(\frac{a^2-1}{a^2})}{4}$ $\endgroup$ – user565198 Jul 20 '18 at 10:06
  • $\begingroup$ @Bonjour. I did not try since I have to go now. Please, could you let me know if you get an answer ? By the way, do you agree with me that splitting the integral does not seem to be a good idea . Cheers. $\endgroup$ – Claude Leibovici Jul 20 '18 at 10:16
  • $\begingroup$ ok, thank you, I agree. See you. $\endgroup$ – user565198 Jul 20 '18 at 10:40

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