Question Number 217761 by mnjuly1970 last updated on 20/Mar/25
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:{prove}\:{that}\:: \\ $$$$ \\ $$$$ \\ $$$$\:\mathrm{I}=\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\pi{x}\right){sin}\left(\mathrm{2}\pi{x}\right){sin}\left(\mathrm{3}\pi{x}\right)}{{x}^{\mathrm{3}} }\:=\:\pi^{\mathrm{3}} \:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$
Answered by Ghisom last updated on 22/Mar/25
![∫_0 ^∞ ((sin πx sin 2πx sin 3πx)/x^3 )dx= =(1/4)∫_0 ^∞ (((sin 2πx)/x^3 )+((sin 4πx)/x^3 )−((sin 6πx)/x^3 ))dx J(n)=(1/4)∫_0 ^∞ ((sin nπx)/x^3 )dx= [t=nπx → dx=(dt/(nπ))] =((n^2 π^2 )/4) ∫_0 ^∞ ((sin t)/t^3 )dt= [by parts] =((n^2 π^2 )/8)(−[((sin t)/t^2 )]_0 ^∞ +∫_0 ^∞ ((cos t)/t^2 )dt)= [by parts] =−((n^2 π^2 )/8)([((tcos t +sin t)/t^2 )]_0 ^∞ +∫_0 ^∞ ((sin t)/t)dt)= [lim_(t→0^+ ) ((tcos t +sin t)/t^2 ) =lim_(t→∞) ((tcos t +sin t)/t^2 ) =0] [we know that ∫_0 ^∞ ((sin t)/t)dt=(π/2)] =−((n^2 π^2 )/8)(0+(π/2))=−((n^2 π^3 )/(16)) ⇒ I=J(2)+J(4)−J(6)=−(π^3 /4)−π^3 +((9π^3 )/4)=π^3](https://www.tinkutara.com/question/Q217825.png)
$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{sin}\:\pi{x}\:\mathrm{sin}\:\mathrm{2}\pi{x}\:\mathrm{sin}\:\mathrm{3}\pi{x}}{{x}^{\mathrm{3}} }{dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\left(\frac{\mathrm{sin}\:\mathrm{2}\pi{x}}{{x}^{\mathrm{3}} }+\frac{\mathrm{sin}\:\mathrm{4}\pi{x}}{{x}^{\mathrm{3}} }−\frac{\mathrm{sin}\:\mathrm{6}\pi{x}}{{x}^{\mathrm{3}} }\right){dx} \\ $$$${J}\left({n}\right)=\frac{\mathrm{1}}{\mathrm{4}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{sin}\:{n}\pi{x}}{{x}^{\mathrm{3}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}={n}\pi{x}\:\rightarrow\:{dx}=\frac{{dt}}{{n}\pi}\right] \\ $$$$=\frac{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }{\mathrm{4}}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{sin}\:{t}}{{t}^{\mathrm{3}} }{dt}= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=\frac{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }{\mathrm{8}}\left(−\left[\frac{\mathrm{sin}\:{t}}{{t}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\infty} +\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{cos}\:{t}}{{t}^{\mathrm{2}} }{dt}\right)= \\ $$$$\:\:\:\:\:\left[\mathrm{by}\:\mathrm{parts}\right] \\ $$$$=−\frac{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }{\mathrm{8}}\left(\left[\frac{{t}\mathrm{cos}\:{t}\:+\mathrm{sin}\:{t}}{{t}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\infty} +\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{sin}\:{t}}{{t}}{dt}\right)= \\ $$$$\:\:\:\:\:\left[\underset{{t}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\:\frac{{t}\mathrm{cos}\:{t}\:+\mathrm{sin}\:{t}}{{t}^{\mathrm{2}} }\:=\underset{{t}\rightarrow\infty} {\mathrm{lim}}\:\frac{{t}\mathrm{cos}\:{t}\:+\mathrm{sin}\:{t}}{{t}^{\mathrm{2}} }\:=\mathrm{0}\right] \\ $$$$\:\:\:\:\:\left[\mathrm{we}\:\mathrm{know}\:\mathrm{that}\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\:\frac{\mathrm{sin}\:{t}}{{t}}{dt}=\frac{\pi}{\mathrm{2}}\right] \\ $$$$=−\frac{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }{\mathrm{8}}\left(\mathrm{0}+\frac{\pi}{\mathrm{2}}\right)=−\frac{{n}^{\mathrm{2}} \pi^{\mathrm{3}} }{\mathrm{16}} \\ $$$$\Rightarrow \\ $$$${I}={J}\left(\mathrm{2}\right)+{J}\left(\mathrm{4}\right)−{J}\left(\mathrm{6}\right)=−\frac{\pi^{\mathrm{3}} }{\mathrm{4}}−\pi^{\mathrm{3}} +\frac{\mathrm{9}\pi^{\mathrm{3}} }{\mathrm{4}}=\pi^{\mathrm{3}} \\ $$
Commented by mnjuly1970 last updated on 22/Mar/25
$${thanks}\:{alot}\:{sir}\:.{very}\:{nice}\:{solution} \\ $$