About the integral $\int_0^{\pi/2} (\sin x-1)/\ln(\sin x) \mathrm{d}x$

Question

Using the Feynman Technique,

Let $$I(a) := \int_0^{\pi/2} \frac{(\sin x)^a - 1}{\ln(\sin x)} \mathrm{d}x$$

$$I’(a)= \int_0^{\pi/2} (\sin x)^a \mathrm{d}x = \frac{\sqrt{π}}{2} \frac{\Gamma\left(\frac{a+1}{2}\right)}{ \Gamma\left(\frac{a+2}{2}\right)}$$

$$I(a)= \frac{\sqrt{π}}{4} \frac{\Gamma\left(\frac{a+1}{2}\right)\left(\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a+2}{2}\right)\right)}{\Gamma\left(\frac{a+2}{2}\right)} +C$$

The constant I’ve got is $\frac{π \ln 2}{2}$.

Therefore,

$$\begin{split} \int_0^{π/2} (\sin x-1)/\ln(\sin x) \mathrm{d}x &= I(1)\\ &= \frac{\sqrt{π}}{4} \frac{\Gamma(1)(\psi(1)-\psi(3/2))}{\Gamma(3/2)} + \frac{\pi \ln 2}{2}\\ &= \frac{1}{2} (2\ln 2 -2) + \frac{\pi \ln 2}{2}\\ &= \ln 2 - 1 + \frac{\pi \ln 2}{2}\\ &\approx 0.782 \end{split}$$

But WolframAlpha gave me the result about $1.23$.

What mistakes did I just make?Thank you

Answer 1

The quantity $$\frac{\sqrt{\pi}}{4} \frac{\Gamma\left(\frac{a+1}{2}\right)\left(\psi\left(\frac{a+1}{2}\right)-\psi\left(\frac{a+2}{2}\right)\right)}{\Gamma\left(\frac{a+2}{2}\right)}$$ is $I''(a)$, not $I(a)$ as you wrote. I don't know how to express $$I(1) - I(0) = \frac{\sqrt{\pi}}{2} \int_0^1 \frac{\Gamma\left(\frac{a+1}{2}\right)}{\Gamma\left(\frac{a+2}{2}\right)} \,da$$ in simpler terms, nor does Mathematica or Maple.

Answer 2

If you want to compute $$I=\int_0^1 \frac{\Gamma\left(\frac{a+1}{2}\right)}{\Gamma\left(\frac{a+2}{2}\right)} \,da$$ you can approximate the integrand by its $P_{n,n}$ Padé approximant built around $a=0$.

Then, for $n \leq 3$, partial fraction decomposition leads to (very nasty) explicit results. Converted to decimals, the numerical valus are

$$\left( \begin{array}{cc} n & \int_0^1 P_{n,n}\, da \\ 1 & \color{red}{1.38}68540 \\ 2 & \color{red}{1.3833}219 \\ 3 & \color{red}{1.3833026} \\ \end{array} \right)$$