Yesterday when I was playing with Wolfram Alpha online calculator I wondered about how to calculate an approximation of this integral $$\int_0^{\pi/2}x\cdot\Re\left((-1+\sin x)^{\log x}\right)dx \tag{1}$$ where $\Re(z)$ is the real part function.

See this code

int x Re((-1+sin x)^(log(x)))dx, from x=0 to pi/2

This is an example of integrals like this $$\int_0^{\pi/2}\left(-1+\sin x\right)^{\log x} dx$$ that I don't know if was in the literature.

How and which we can find an approximation (similar than that provide us Wolfram Alpha) of the integral $(1)$?

I don't know if we can get such approximation using calculus, or well is required using real analysis or complex analysis.

Question. Can you provide us an approximation, or references for, of $$\int_0^{\pi/2}x\cdot\Re\left((-1+\sin x)^{\log x}\right)dx,$$ justifying your calculations? Many thanks.

  • 1
    $\begingroup$ A suitable approximation for the highlighted part of the graph in WolframAlpha may be $\cos7x+0.4$ since it cuts the axis at around $0.28$ and $0.61$, and also has a local (global) minimum of around $-0.6$. So the integral is $$\int_{0.28}^{0.61}\cos7x+0.4\, dx=\left[\frac17\sin7x+0.4x\right]_{0.28}^{0.61}\approx-0.129$$ This is similar to wolframalpha.com/input/…, with an approximate value of around $-0.147$. $\endgroup$ – TheSimpliFire Dec 16 '17 at 17:30
  • $\begingroup$ Many thanks for your calculations and edit, feel free to add your contribution as an answer @TheSimpliFire $\endgroup$ – user243301 Dec 16 '17 at 21:22

From the graph of WolframAlpha, we can split the area into four sections: from $0.08$ to $0.22$, $0.22$ to $0.6$, $0.6$ to $1$ and $1$ to $\dfrac\pi2$.

These approximations can be visualized using Desmos.

Cubic equation seems suitable:$$I_1=\int_{0.08}^{0.22}(x-0.08)(x-0.22)(x-40)\, dx=0.0182$$ Function involving cosine seems suitable (adjust limits): $$I_2 = \int_{0.25}^{0.64}\cos7x+0.2\, dx=-0.2016$$ Cubic equation seems suitable: $$I_3=\int_{0.6}^{1}(x-0.6)(x-1.4)(x-7)\, dx=0.2624$$ Function involving cosine seems suitable: $$I_4 = \int_{1}^{\pi/2}\cos(\pi x-\pi)\, dx=0.3105$$ Total area $\approx I_1 + I_2 + I_3 + I_4=0.3895$ which is a reasonable estimate to the actual value of $0.40002$.

  • $\begingroup$ Many thanks I am going to study (check) your method and estimation. $\endgroup$ – user243301 Dec 23 '17 at 10:41

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