Question Number 219388 by SdC355 last updated on 23/Apr/25
![∫_0 ^( ∞) ((sin^2 (u))/u^2 ) du=I I(t)=∫_0 ^( ∞) ((sin^2 (u))/u^2 )e^(−ut) du=∫_0 ^( ∞) (1/u)∙((sin^2 (u))/u)e^(−ut) du= ∫_( t) ^( ∞) L_u {((sin^2 (p))/p)} du=∫_( t) ^( ∞) ∫_0 ^( ∞) ((sin^2 (p))/p)e^(−up) dpdu ∫_( t) ^( ∞) ∫_( w) ^( ∞) L_s {sin^2 (r)}ds dw ∫_( t) ^( ∞) ∫_( w) ^( ∞) ∫_0 ^( ∞) sin^2 (r)e^(−sr) dr ds dw ∫_( t) ^( ∞) ∫_( w) ^( ∞) (2/(s(s^2 +4))) ds dw=∫_( t) ^( ∞) [(1/2)ln(s)−(1/4)ln(s^2 +4)]_(s=w) ^(s=∞) dw ∫_( t) ^( ∞) (1/4)ln((w^2 /(w^2 +4)))dw=[(1/4)w∙ln(1+((2/w))^2 )+tan^(−1) ((1/2)w)]_(w=t) ^(w=∞) (π/2)−(1/4)t∙ln(1+((2/t))^2 )−tan^(−1) ((1/2)t) lim_(t→0) {tan^(−1) ((2/t))−(1/4)t∙ln(1+((2/t))^2 )}=(π/2)](https://www.tinkutara.com/question/Q219388.png)
$$\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}^{\mathrm{2}} }\:\mathrm{d}{u}={I} \\ $$$${I}\left({t}\right)=\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}^{\mathrm{2}} }{e}^{−{ut}} \mathrm{d}{u}=\int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{1}}{{u}}\centerdot\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}}{e}^{−{ut}} \mathrm{d}{u}= \\ $$$$\int_{\:{t}} ^{\:\infty} \:\:\mathcal{L}_{{u}} \left\{\frac{\mathrm{sin}^{\mathrm{2}} \left({p}\right)}{{p}}\right\}\:\mathrm{d}{u}=\int_{\:{t}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({p}\right)}{{p}}{e}^{−{up}} \mathrm{d}{p}\mathrm{d}{u} \\ $$$$\int_{\:{t}} ^{\:\infty} \:\int_{\:{w}} ^{\:\infty} \:\mathcal{L}_{{s}} \left\{\mathrm{sin}^{\mathrm{2}} \left({r}\right)\right\}\mathrm{d}{s}\:\mathrm{d}{w} \\ $$$$\int_{\:{t}} ^{\:\infty} \int_{\:{w}} ^{\:\infty} \int_{\mathrm{0}} ^{\:\infty} \:\mathrm{sin}^{\mathrm{2}} \left({r}\right){e}^{−{sr}} \mathrm{d}{r}\:\mathrm{d}{s}\:\mathrm{d}{w} \\ $$$$\int_{\:{t}} ^{\:\infty} \int_{\:{w}} ^{\:\infty} \:\:\frac{\mathrm{2}}{{s}\left({s}^{\mathrm{2}} +\mathrm{4}\right)}\:\mathrm{d}{s}\:\mathrm{d}{w}=\int_{\:{t}} ^{\:\infty} \:\:\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({s}\right)−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left({s}^{\mathrm{2}} +\mathrm{4}\right)\right]_{{s}={w}} ^{{s}=\infty} \mathrm{d}{w} \\ $$$$\int_{\:{t}} ^{\:\infty} \:\:\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\left(\frac{{w}^{\mathrm{2}} }{{w}^{\mathrm{2}} +\mathrm{4}}\right)\mathrm{d}{w}=\left[\frac{\mathrm{1}}{\mathrm{4}}{w}\centerdot\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{w}}\right)^{\mathrm{2}} \right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}{w}\right)\right]_{{w}={t}} ^{{w}=\infty} \\ $$$$\frac{\pi}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}{t}\centerdot\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{t}}\right)^{\mathrm{2}} \right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{1}}{\mathrm{2}}{t}\right) \\ $$$$\underset{{t}\rightarrow\mathrm{0}} {\mathrm{lim}}\left\{\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{2}}{{t}}\right)−\frac{\mathrm{1}}{\mathrm{4}}{t}\centerdot\mathrm{ln}\left(\mathrm{1}+\left(\frac{\mathrm{2}}{{t}}\right)^{\mathrm{2}} \right)\right\}=\frac{\pi}{\mathrm{2}} \\ $$
Answered by SdC355 last updated on 23/Apr/25
$$\mathrm{But}\:\mathrm{why}\:\mathrm{am}\:\mathrm{i}\:\mathrm{wrong}???\: \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{sheet}\:\mathrm{saying}\:\int_{\mathrm{0}} ^{\:\infty} \:\:\frac{\mathrm{sin}^{\mathrm{2}} \left({u}\right)}{{u}^{\mathrm{2}} }\:\mathrm{d}{u}\:\mathrm{is}\:\frac{\pi}{\mathrm{4}}.. \\ $$
Commented by vnm last updated on 23/Apr/25
$$ \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin}^{\mathrm{2}} {u}}{{u}^{\mathrm{2}} }\mathrm{d}{u}=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{d}}{\mathrm{d}{u}}\left(−\frac{\mathrm{1}}{{u}}\right)\mathrm{sin}^{\mathrm{2}} {u}\mathrm{d}{u}= \\ $$$$−\frac{\mathrm{1}}{{u}}\mathrm{sin}^{\mathrm{2}} {u}\mid_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} −\frac{\mathrm{1}}{{u}}\frac{\mathrm{d}}{\mathrm{d}{u}}\left(\mathrm{sin}^{\mathrm{2}} {u}\right)\mathrm{d}{u}= \\ $$$$\mathrm{0}+\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{sin2}{u}}{\mathrm{2}{u}}\mathrm{d}\left(\mathrm{2}{u}\right)=\frac{\pi}{\mathrm{2}} \\ $$