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I was not sure how to go about finding a closed form for $$\int\log\left(\cos\left(\frac{a}{x}\right)\right) \ dx\,\,\,\,(1)$$

so I was wondering if there is some clever integration trick that would work?

(Perhaps using $a$ as an integral upper limit and lower limit $>=0$ might produce some result as a definite integral?)

Some work done so far by me...

Trying an asymptotic expansion using a integration by parts with a=1 seems to look like things will get rather complicated quick or at least I cannot see an easily identifieble series emerging for (1) How can one tell in advance? $$-\frac{1}{2} x^2 \left(-\frac{2 \tan ^2\left(\frac{1}{x}\right)}{x^3 \log ^3\left(\cos \left(\frac{1}{x}\right)\right)}-\frac{\sec ^2\left(\frac{1}{x}\right)}{x^3 \log ^2\left(\cos \left(\frac{1}{x}\right)\right)}-\frac{\tan \left(\frac{1}{x}\right)}{x^2 \log ^2\left(\cos \left(\frac{1}{x}\right)\right)}\right)+\frac{x}{\log \left(\cos \left(\frac{1}{x}\right)\right)}+\frac{\tan \left(\frac{1}{x}\right)}{\log ^2\left(\cos \left(\frac{1}{x}\right)\right)}+\frac{\tan \left(\frac{1}{x}\right)}{x \log ^2\left(\cos \left(\frac{1}{x}\right)\right)}$$

onepound
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    What is the original problem you're after? Wasn't there a definite integral or something? – Zacky Dec 29 '19 at 20:32
  • @zacky After looking at $\int_0^{\pi/2}\log\cos(x),\mathrm{d}x$ https://math.stackexchange.com/questions/690644/evaluate-int-0-pi-2-log-cosx-mathrmdx I wondered if 1/x had been investigated over some range that is why I left any limit out. Also I've removed the second question because it does not help and is a bit of a whale sandwich. – onepound Dec 30 '19 at 10:11
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    Well, we have:$$\int \ln\left(\cos\left(\frac{a}{x}\right)\right)dx\overset{\frac{a}{x}=t}=-a\int \frac{\ln \cos t}{t^2}dt\overset{IBP}=\frac{a}t\ln \cos t +a\int \frac{\tan t}{t}dt$$ I suggest to not waste too much effort around it without a specific motivation. – Zacky Dec 30 '19 at 10:39
  • @zacky okay thank you for reworking it into $\frac{a}t\ln \cos t +a\int \frac{\tan t}{t}dt$. – onepound Dec 30 '19 at 13:26

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