Prove that for $a,b,c>0$ $$\int_0^\infty \frac{|\sin (ax)|^c- |\sin(bx)|^c}{x} \, dx = \log\left(\frac{a}{b}\right) \frac{\Gamma(1+c)}{2^c\Gamma^2\left(1+\frac c2\right)}.$$
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2$\begingroup$ I'm afraid the answer is not correct - I checked numerically for $c=\frac{1}{2}; a=2; b=1$. Using the Frullani' integral for a periodic function, and defining $\,0<c<1; \,\,a,b>0$, the correct answer is $$ \int_0^\infty \frac{|\sin (ax)|^c- |\sin(bx)|^c}{x} \, dx = \log\left(\frac{a}{b}\right) \frac{\Gamma\left(c+1\right)}{2^c\,\Gamma^2\left(1+\frac c2\right)} $$ what gives $0.528...$ for both sides at $c=\frac{1}{2}; a=2; b=1$ $\endgroup$– SvyatoslavJun 15, 2022 at 8:02
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2$\begingroup$ Unfortunately for your strategy, $\Im e^{iacx}=\sin(acx)\not\equiv|\sin(ax)|^c$. $\endgroup$– J.G.Jun 15, 2022 at 8:38
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1$\begingroup$ Also, very useful more general relation $$\int_0^\pi e^{ibx}\sin^{a-1}x~dx=\frac{\pi e^{i\pi b/2}}{2^{a-1}a\mathrm{B}\left(\frac{a+b+1}{2},\frac{a-b+1}{2}\right)}$$ A comprehensive proof may be found here: math.stackexchange.com/questions/3419834/… $\endgroup$– SvyatoslavJun 15, 2022 at 9:08
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3$\begingroup$ Also, could be useful pnas.org/doi/pdf/10.1073/pnas.35.10.612 $\endgroup$– SvyatoslavJun 15, 2022 at 12:22
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1$\begingroup$ @Svyatoslav thanks for taking your time! The second link is useful. $\endgroup$– DhanvinJun 15, 2022 at 12:43
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The "periodic Frullani" approach (that I use in this answer): for any $\tau$-periodic function $f$ such that the integral $\int_0^\tau\big(f(x)/x\big)\,dx$ exists (in Lebesgue's sense), and any $a,b>0$, we have $$\int_0^\infty\frac{f(ax)-f(bx)}{x}\,dx=\frac1\tau\log\frac{a}{b}\int_0^\tau f(x)\,dx.$$ Take $f(x)=|\sin x|^c$ (and $\tau=\pi$) and use $\int_0^\pi(\sin x)^c\,dx=\mathrm{B}\big(1/2,(c+1)/2\big)$.
answered Jun 15, 2022 at 8:53