Evaluate : ∫ 0 π 2 ln 2 ⁡ ( cos 2 ⁡ x ) d x

Question

Evaluate :      ∫    0                  π        2                  ln    2    ⁡      (          cos      2        ⁡    x    )    d  x

Steinherrjm · Accepted Answer

I find a way to get the number using gamma functions, nothing is rigorous.
Consider the integral

I (β) = \int_{- \frac{π}{2}}^{\frac{π}{2}} (\cos x)^{β} d x

. We know:

2 \frac{d^{2}}{d β^{2}} I (β) |_{β = 0} = 4 \int_{0}^{\frac{π}{2}} \ln^{2} (\cos x) d x

is the integral we want. Introduce

u = \frac{1 + \sin x}{2}

, we have:

\begin{aligned} I (β) & = \int_{- \frac{π}{2}}^{\frac{π}{2}} (\cos x)^{β - 1} d \sin x \\ = \int_{0}^{1} (4 u (1 - u))^{\frac{β - 1}{2}} d (2 u) \\ = 2^{β} \int_{0}^{1} u^{\frac{β + 1}{2} - 1} (1 - u)^{\frac{β + 1}{2} - 1} d u \\ = 2^{β} \frac{Γ (\frac{β + 1}{2})^{2}}{Γ (β + 1)} \end{aligned}

Using the taylor expansion of various terms at

β = 0

\begin{aligned} 2^{β} & = 1 + \ln (2) β + \frac{\ln^{2} 2}{2} β^{2} + . . . \\ Γ (\frac{β + 1}{2}) & = \sqrt{π} (1 - \frac{γ + 2 \ln 2}{2} β + \frac{π^{2} + 2 (γ + 2 \ln 2)^{2}}{16} β^{2} + . . .) \\ Γ (β + 1) & = 1 - γ β + \frac{6 γ^{2} + π^{2}}{12} β^{2} + . . . \end{aligned}

We get:

\begin{aligned} I (β) = π (1 - \ln (2) β + \frac{π^{2} + 12 \ln^{2} 2}{24} β^{2} + . . .) \\ ⟹ & 2 \frac{d^{2}}{d β^{2}} I (β) |_{β = 0} = 2 π (\frac{π^{2}}{12} + \ln^{2} 2) \end{aligned}

Jaylyn Horne · Accepted Answer

∫          0                      π        2                                sin                    2        m        −        1                    (      x      )                      cos                    2        n        −        1                    (      x      )        d    x    =  B      (    m    ,    n    )          ∫          0                      π        2                                sin                    2        m        −        1                    (      x      )                                (                                    cos                                      2                                x          )                                                  2            n            −            1                    2                      d    x    =  B      (    m    ,    n    )  On differentiating it twice w.r.t. to n and taking   m  =      1    2   and   n  =      1    2  , we get      ∫          0                      π        2                  (    ln    ⁡                  (                              cos                                2                          x        )            d      x              )      2        =      1    2                                (          Γ                      (                          1              2                        )                    )                            2              1        {                            (          ψ                      (                          1              2                        )                    −          ψ                      (            1            )                    )                            2              +          ψ      ′              (              1        2            )        −          ψ      ′              (      1      )        }    ∴      ∫          0                      π        2                  (    ln    ⁡                  (                              cos                                2                          x        )            d      x              )      2        =                    π                    3              6    +  2  π                    (        ln        ⁡                  2                )                    2

Evaluate : int_0^(pi/2) ln^2(cos^2 x)dx

Answered question

Answer & Explanation

New Questions in Calculus 1