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I am trying to find a general expression for $$ \int_0^{\pi/2} (\ln(\cos(x)))^n dx $$ for integer $n$, but have not been able to find it online or derive it. I have a good idea that the general form will involve terms with $\zeta(n)$, but little else.

For reference, here are the first few values for $n$ = 1,2,3,4: $$\int_0^{\pi/2} (\ln(\cos(x))) dx =-\pi\ln(2) $$

$$\int_0^{\pi/2} (\ln(\cos(x)))^2 dx = \frac{\pi}{2} \ln(2)^2 + \frac{\pi^3}{24} $$ $$ = \frac{\pi}{2} \ln(2)^2 + \frac{\pi}{4} \zeta(2) $$

$$\int_0^{\pi/2} (\ln(\cos(x)))^3 dx = - \frac{\pi}{2} \ln(2)^3 -\frac{\pi^3}{8} \ln(2)- \frac{3 \pi}{4} \zeta(3) $$ $$ = - \frac{\pi}{2} \ln(2)^3 -\frac{3\pi}{4}\zeta(2) \ln(2)- \frac{3 \pi}{4} \zeta(3) $$

$$\int_0^{\pi/2} (\ln(\cos(x)))^4 dx = \frac{\pi}{2} \ln(2)^4 + \frac{\pi^3}{4} \ln(2)^2 + {3\pi}\zeta(3)\ln(2) +\frac{19 \pi^5}{480}$$ $$= \frac{\pi}{2}\ln(2)^4 + \frac{3\pi}{2}\zeta(2) \ln(2)^2 + {3\pi}\zeta(3)\ln(2) + \frac{57\pi}{16}\zeta(4)$$

I would enormously appreciate some help. Thank you!

Ron Shvartsman
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1 Answers1

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Recall that $$\int_{0}^{\frac{\pi}{2}} \sin^p(x) \cos^q(x) \: dx = \frac{ \Gamma(\frac{p+1}{2}) \Gamma(\frac{q+1}{2}) }{2 \Gamma(\frac{p+q}{2}+1)} \\ \frac{d^n}{dq^n} \cos^q(x) = \cos^q(x)\ln^n(\cos(x)) $$

so that $$ \int_{0}^{\frac{\pi}{2}} \cos^q(x)\ln^n(\cos(x)) \: dx = \frac{d^n}{dq^n} \frac{ \Gamma(\frac{1}{2})\Gamma(\frac{q+1}{2}) }{2 \Gamma(\frac{q}{2}+1)} \\ \int_{0}^{\frac{\pi}{2}} \ln^n(\cos(x)) \: dx = \lim_{q \to 0}\frac{d^n}{dq^n} \frac{ \Gamma(\frac{1}{2})\Gamma(\frac{q+1}{2}) }{2 \Gamma(\frac{q}{2}+1)}$$ The zeta functions come from the fact that $\psi^{(s)}(1) = \zeta(s+1)(-1)^{s+1} s! $ for $s >0$

DecarbonatedOdes
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