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Prove that for $a,b,c>0$ $$\int_0^\infty \frac{|\sin (ax)|^c- |\sin(bx)|^c}{x} \, dx = \log\left(\frac{a}{b}\right) \frac{\Gamma(1+c)}{2^c\Gamma^2\left(1+\frac c2\right)}.$$

Dhanvin
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    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 $c=\frac{1}{2}; a=2; b=1$ – Svyatoslav Jun 15 '22 at 08:02
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    Unfortunately for your strategy, $\Im e^{iacx}=\sin(acx)\not\equiv|\sin(ax)|^c$. – J.G. Jun 15 '22 at 08:38
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    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: https://math.stackexchange.com/questions/3419834/integrals-related-to-the-reciprocal-beta-function?noredirect=1 – Svyatoslav Jun 15 '22 at 09:08
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    Also, could be useful https://www.pnas.org/doi/pdf/10.1073/pnas.35.10.612 – Svyatoslav Jun 15 '22 at 12:22
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    @Svyatoslav thanks for taking your time! The second link is useful. – Dhanvin Jun 15 '22 at 12:43

2 Answers2

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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 $\int_0^\pi(\sin x)^c\,dx=\mathrm{B}\big(1/2,(c+1)/2\big)$.

metamorphy
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  • I think you are missing the term near $0$: $\log(a/b)\Big(\bar{f}-f(0+)\Big)$ where $\bar{f}=\frac1\tau\int^\tau_0f$ – Mittens May 25 '23 at 00:35
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    No need for $\int^1_0\frac{f(x)}{x},dx$ as you can see in the proof that posted. The possible singularity at $0$ is controlled by the difference $f(ax)-f(bx)$ with the additional conditions I laid down. – Mittens May 25 '23 at 03:00
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This is just to elaborate on my comment to @matamorphy's answer.

Assume $f$ is bounded, $T$-periodic, integrable in $[0,T]$, and that $\lim_{t\rightarrow0+}f(t)=f_0$ exists. For any $0<a<b$, \begin{align} \int^\infty_0\frac{f(at) - f(bt)}{t}\,dt=\log(a/b)\Big(\frac{1}{T}\int^T_0f(t)\,dt - f_0\Big) \end{align} where the integral on the left-hand-side is considered as improper Riemann integral.

Here is a sketch of the proof:
\begin{align} \int^A_\varepsilon\frac{f(at)}{t}\,dt-\int^A_\varepsilon\frac{f(bt)}{t}\,dt&=\int^{aA}_{a\varepsilon} \frac{f(t)}{t}\,dt-\int^{bA}_{b\varepsilon} \frac{f(t)}{t}\,dt\\ &=\int^{b\varepsilon}_{a\varepsilon}\frac{f(t)}{t}\,dt-\int^{Ab}_{Aa}\frac{f(t)}{t}\,dt\\ &=\int^b_a\frac{f(\varepsilon t)}{t}\,dt-\int^a_b\frac{f(At)}{t}\,dt \end{align}

The right continuity of $f$ at $0$ implies that \begin{align}\int^b_a\frac{f(\varepsilon t)}{t}\,dt&\xrightarrow{\varepsilon\rightarrow0}f_0\log(b/a) \end{align} Fejer's Lemma implies that \begin{align}\int^b_a\frac{f(A t)}{t}\,dt&\xrightarrow{A\rightarrow\infty}\log(b/a)\frac{1}{T}\int^T_0f(t)\,dt \end{align} Putting things together yields Frullani's integral formula for periodic functions: \begin{align} \int^\infty_0\frac{f(at) - f(bt)}{t}\,dt=\log(a/b)\Big(\frac{1}{T}\int^T_0f(t)\,dt - f_0\Big) \end{align}

Mittens
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