How can you evaluate \int_0^{\pi/2}\log\cos(x)dx ?

Question

How can you evaluate \int_0^{\pi/2}\log\cos(x)dx ?

kuhse4461a 2021-12-30 Answered

How can you evaluate

\int_{0}^{\frac{π}{2}} \log \cos (x) d x

?

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Expert Answer

Kirsten Davis

Answered 2021-12-31 Author has 27 answers

For the sake of simplicity, all the integral variables I use are x even there are a lot of substitutions. Because lots of variables could make one confused.
Let I denote the integral value. By substitute x for

\frac{π}{2} - x

, we have:

I = \int_{0}^{\frac{π}{2}} \log \cos (x) d x = \int_{0}^{\frac{π}{2}} \log \sin (x) d x

And then, we have:

I = \int_{0}^{\frac{π}{2}} \log (2 \cos (\frac{x}{2}) \sin (\frac{x}{2})) d x

= \frac{π}{2} \log 2 + \int_{0}^{\frac{π}{2}} \log \cos (\frac{x}{2}) d x + \int_{0}^{\frac{π}{2}} \log \sin (\frac{x}{2}) d x

= \frac{π}{2} + 2 \int_{0}^{\frac{π}{4}} \log \cos (x) d x + 2 \int_{0}^{\frac{π}{4}} + 2 \int_{0}^{\frac{π}{4}} \log \sin (x) d x

= \frac{π}{2} \log 2 + I_{1} + I_{2}

In the second step from bottom, I use the substitution that

x = \frac{x}{2}

For

I_{1}

, use the substitution that

x = \frac{π}{2} - x

we obtain

I_{1} = 2 \int_{\frac{π}{4}}^{\frac{π}{2}} \log \sin (x) d x

It gives that

I_{1} + I_{2} = 2 I

. So we have

I = \frac{π}{2} \log 2 + 2 I

I = - \frac{π}{2} \log 2

This is helpful 0

Shawn Kim

Answered 2022-01-01 Author has 25 answers

\int_{0}^{\frac{π}{2}} \ln \cos x d x = I = \int_{0}^{\frac{π}{2}} \ln \sin x d x

By symmetry we have

\ln \cos x = \ln \sin x

on the interval

[0, π .2]

. This is true for any even/odd function on this interval, as is an exercise in Demidovich-Problems in Analysis. Thus we have

2 I = \int_{0}^{\frac{π}{2}} \ln \cos x d x + \int_{0}^{\frac{π}{2}} \ln \sin x d x = \int_{0}^{\frac{π}{2} \ln (\sin x \cos x) d x = \int_{0}^{\frac{π}{2}} \ln (\frac{1}{2} \cdot \sin (2 x)) d x}

All I used was

\ln (a \cdot b) = \ln (a) + \ln (b)

and

2 \sin x \cos x = \sin (2 x)

. Now we split the integral back up to obtain

- \int_{0}^{\frac{π}{2}} \ln (2) d x + \int_{0}^{\frac{π}{2}} \ln (\sin (2 x)) d x = 2 I

But the integral of

\ln \sin u

is

2 I

, thus we have

- \frac{π \ln (2)}{2} + I = 2 I \to I = - \frac{π \ln (2)}{2}

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Vasquez

Answered 2022-01-09 Author has 460 answers

We have a well-known identity:
$\prod_{k = 1}^{n - 1} \sin (\frac{π k}{n}) = \frac{2 n}{2^{n}}$
and since $\log \sin x$ is an improperly Riemann-integrable function over $(0, π)$ , it follows that:
$\int_{0}^{π} \log \sin θ d θ = lim_{n \to + \infty} \frac{π}{n} \sum_{k = 1}^{n - 1} \log \sin (\frac{π k}{n}) = - π \log 2$
so:
$\int_{0}^{π / 2} \log \sin θ d θ = \int_{0}^{π / 2} \log \sin θ d θ = - \frac{π}{2} \log 2$