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By considering the integral \begin{align} I_{\mu} = \int_{0}^{\pi/4} \sin(2\theta) \, \left( \cos(\theta) - \sin(\theta) \right)^{\mu} \, d\theta \end{align} derivatives can be taken with respect to $\mu$ to obtain integrals involving logarithms. Let these integrals be \begin{align} J_{\mu}^{m} = \partial_{\mu}^{m} I_{\mu} = \int_{0}^{\pi/4} \sin(2\theta) \, \left( \cos(\theta) - \sin(\theta) \right)^{\mu} \, \ln^{m}(\cos\theta - \sin\theta) \, d\theta. \end{align}

What are the closed form values of $I_{\mu}$, $J_{0}^{0}$, $J_{0}^{1}$, and $J_{0}^{2}$ ?

Is it possible to also to extend the results to the integral \begin{align} T_{\mu}^{k} = \int_{0}^{\pi/4} \sin(2\theta) \left( \cos^{2/k}\theta - \sin^{2/k}\theta \right)^{\mu} \, d\theta \hspace{3mm} ? \end{align}

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2 Answers 2

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Another possible closed-form

$$I_\mu=\frac{2\cdot{_2F_1}\left(\begin{array}c\tfrac12,2\\\tfrac{\mu+5}{2}\end{array}\middle|\,-1\right)}{(\mu+3)(\mu+1)},$$

for $\Re(\mu)>-1$.

I don't know how to simplify it further, but I think it's a good idea for integer $\mu$ values to separate the even and the odd $\mu$ cases, since

$$\begin{align} I_0 & = \frac 12\\ I_1 & = \frac 23 - \frac{\sqrt{2}}{3}\\ I_2 & = \frac 12 - \frac{\pi}{8}\\ I_3 & = \frac 65 - \frac{\sqrt{2}}{5}\\ I_4 & = \frac 56 - \frac{\pi}{4}\\ I_5 & = \frac{46}{21} - \frac{32 \sqrt{2}}{21}\\ I_6 & = \frac 32 - \frac{15\pi}{32}\\ I_7 & = \frac{182}{45} - \frac{128\sqrt{2}}{45}\\ I_8 & = \frac{83}{30} - \frac{7\pi}{8}\\ I_9 & = \frac{2902}{385} - \frac{2048\sqrt{2}}{385}. \end{align}$$

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May be, you could be interested by the fact that $$\begin{align} I_{\mu} = \int_{0}^{\pi/4} \sin(2\theta) \, \left( \cos(\theta) - \sin(\theta) \right)^{\mu} \, d\theta=\frac{\, _2F_1\left(-\frac{1}{2},1;\frac{\mu+3}{2};-1\right)-\, _2F_1\left(\frac{1}{2},1;\frac{\mu+3}{2};-1\right)}{\mu+1} \end{align}$$ provided that $\Re(\mu)>-1$

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