Question Number 210554 by Batmath last updated on 12/Aug/24
Answered by MrGaster last updated on 02/Nov/24
![∫_0 ^∞ [(1/2)−S(px)]x^(2q−1) dx=∫_0 ^∞ [(1/2)−(1/π)∫_0 ^∞ ((sin(tpx))/t)dt]x^(2q−1) dx =(1/π)∫_0 ^∞ ∫_0 ^∞ (1/2)[(1/2)−((sin(tpx))/t)]x^(2q−1) dtdx =(1/π)∫_0 ^∞ ∫_0 ^∞ (1/2)x^(2q−1) dxdt−(1/π)∫_0 ^∞ ∫_0 ^∞ ((sin(tpx))/t)x^(2q−1) dtdx =(1/π)∫_0 ^∞ (1/2) (x^(2q) /(2q))∣_0 ^∞ dt−(1/π)∫_0 ^∞ (1/t)[((−cos(tpx))/(2qp))x^(2q) ]∣_0 ^∞ dt =(1/π)∫_0 ^∞ (1/2)∙(1/(2a))lim_(x→∞) x^(2a) dt−(1/x)∫_0 ^∞ (1/t)∙(1/(2qp))[lim_(x→∞) cos(tqx)x^(2q) −lim_(x→0) cos(tqx)x^(2q) ]dt =(1/π)∙(1/(2q))∙(1/2)∙∞−(1/π)∙(1/(2qp))[0−1]∫_0 ^∞ (1/t^(2q+1) )dt =(1/π)∙(1/(2qp))∙(π/2)∙(1/(Γ(2q+1))) =(1/(4qp))∙((Γ(2q+1))/(Γ(2q+1))) =(1/(4qp))∙((Γ(2q+1))/(2qΓ(2q))) =(1/(4qp))∙((Γ(2q+1))/(2(√π)Γ(2q+(1/2)))) =(√2)((Γ(q+(1/2))sin((2q+1)/4)π)/(4(√π)qp^(2q) ))](https://www.tinkutara.com/question/Q213259.png)
$$\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{1}}{\mathrm{2}}−{S}\left({px}\right)\right]{x}^{\mathrm{2}{q}−\mathrm{1}} {dx}=\int_{\mathrm{0}} ^{\infty} \left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left({tpx}\right)}{{t}}{dt}\right]{x}^{\mathrm{2}{q}−\mathrm{1}} {dx} \\ $$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{sin}\left({tpx}\right)}{{t}}\right]{x}^{\mathrm{2}{q}−\mathrm{1}} {dtdx} \\ $$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}{q}−\mathrm{1}} {dxdt}−\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}\left({tpx}\right)}{{t}}{x}^{\mathrm{2}{q}−\mathrm{1}} {dtdx} \\ $$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}}\:\frac{{x}^{\mathrm{2}{q}} }{\mathrm{2}{q}}\mid_{\mathrm{0}} ^{\infty} {dt}−\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}}\left[\frac{−\mathrm{cos}\left({tpx}\right)}{\mathrm{2}{qp}}{x}^{\mathrm{2}{q}} \right]\mid_{\mathrm{0}} ^{\infty} {dt} \\ $$$$=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\mathrm{2}{a}}\underset{{x}\rightarrow\infty} {\mathrm{lim}}{x}^{\mathrm{2}{a}} {dt}−\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}}\centerdot\frac{\mathrm{1}}{\mathrm{2}{qp}}\left[\underset{{x}\rightarrow\infty} {\mathrm{lim}cos}\left({tqx}\right){x}^{\mathrm{2}{q}} −\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}cos}\left({tqx}\right){x}^{\mathrm{2}{q}} \right]{dt} \\ $$$$=\frac{\mathrm{1}}{\pi}\centerdot\frac{\mathrm{1}}{\mathrm{2}{q}}\centerdot\frac{\mathrm{1}}{\mathrm{2}}\centerdot\infty−\frac{\mathrm{1}}{\pi}\centerdot\frac{\mathrm{1}}{\mathrm{2}{qp}}\left[\mathrm{0}−\mathrm{1}\right]\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{t}^{\mathrm{2}{q}+\mathrm{1}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\pi}\centerdot\frac{\mathrm{1}}{\mathrm{2}{qp}}\centerdot\frac{\pi}{\mathrm{2}}\centerdot\frac{\mathrm{1}}{\Gamma\left(\mathrm{2}{q}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{qp}}\centerdot\frac{\Gamma\left(\mathrm{2}{q}+\mathrm{1}\right)}{\Gamma\left(\mathrm{2}{q}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{qp}}\centerdot\frac{\Gamma\left(\mathrm{2}{q}+\mathrm{1}\right)}{\mathrm{2}{q}\Gamma\left(\mathrm{2}{q}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{qp}}\centerdot\frac{\Gamma\left(\mathrm{2}{q}+\mathrm{1}\right)}{\mathrm{2}\sqrt{\pi}\Gamma\left(\mathrm{2}{q}+\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\sqrt{\mathrm{2}}\frac{\Gamma\left({q}+\frac{\mathrm{1}}{\mathrm{2}}\right)\mathrm{sin}\frac{\mathrm{2}{q}+\mathrm{1}}{\mathrm{4}}\pi}{\mathrm{4}\sqrt{\pi}{qp}^{\mathrm{2}{q}} } \\ $$