Question Number 203564 by mnjuly1970 last updated on 22/Jan/24
$$ \\ $$$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\:\mathrm{3}} \left({x}\right)}{{x}^{\:\mathrm{2}} }\:{dx}=\:?\:\:\:\:\: \\ $$
Answered by Mathspace last updated on 23/Jan/24
![I=[−(1/x)sin^3 x]_0 ^∞ −∫_0 ^∞ (−(1/x))3cosx sin^2 x dx =3∫_0 ^∞ ((cosx sin^2 x)/x)dx =3∫_0 ^∞ ((cosx(1−cos(2x)))/(2x))dx =(3/2)∫_0 ^∞ ((cosx−cosxcos(2x))/x)dx cosx cos(2x)=(1/2)(cos(3x)+cosx) ⇒I=(3/2)∫_0 ^∞ ((cosx−(1/2)cos(3x)−(1/2)cosx)/x)dx =(3/4)∫_0 ^∞ ((cosx−cos(3x))/x)dx f(a)=∫_0 ^∞ (((cosx−cos(3x))e^(−ax) )/x)dx f^′ (a)=−∫_0 ^∞ (cosx)e^(−ax) −cos(3x)e^(−ax) )dx =∫_0 ^∞ cos(3x)e^(−ax) dx−∫_0 ^∞ cosx e^(−ax) dx =Re(∫_0 ^∞ e^(3ix−ax) dx)−Re(∫_0 ^∞ e^(ix−ax) dx) but ∫_0 ^∞ e^((−a+3i)x) dx =(1/(−a+3i))e^((−a+3i)) ]_0 ^∞ =(1/(a−3i)) =((a+3i)/(a^2 +9)) and ∫_0 ^∞ e^((−a+i)x) dx=[(1/(−a+i))e^((−a+i)x) ]_0 ^∞ =(1/(a−i))=((a+i)/(a^2 +1)) ⇒f^′ (a)=(a/(a^2 +9))−(a/(a^2 +1)) ⇒f(a)=(1/2)ln(((a^2 +9)/(a^2 +1)))+k k=lim_(a→∞) f(a)=0 ⇒ f(a)=(1/2)ln(((a^2 +9)/(a^2 +1))) ∫_0 ^∞ ((cosx−cos(3x))/x)dx=f(0)=(1/2)ln9 =ln3 ⇒I=(3/4)ln3](https://www.tinkutara.com/question/Q203628.png)
$${I}=\left[−\frac{\mathrm{1}}{{x}}{sin}^{\mathrm{3}} {x}\right]_{\mathrm{0}} ^{\infty} −\int_{\mathrm{0}} ^{\infty} \left(−\frac{\mathrm{1}}{{x}}\right)\mathrm{3}{cosx}\:{sin}^{\mathrm{2}} {x}\:{dx} \\ $$$$=\mathrm{3}\int_{\mathrm{0}} ^{\infty} \:\frac{{cosx}\:{sin}^{\mathrm{2}} {x}}{{x}}{dx} \\ $$$$=\mathrm{3}\int_{\mathrm{0}} ^{\infty} \frac{{cosx}\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right)}{\mathrm{2}{x}}{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{cosx}−{cosxcos}\left(\mathrm{2}{x}\right)}{{x}}{dx} \\ $$$${cosx}\:{cos}\left(\mathrm{2}{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\left(\mathrm{3}{x}\right)+{cosx}\right) \\ $$$$\Rightarrow{I}=\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{cosx}−\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{3}{x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{cosx}}{{x}}{dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{cosx}−{cos}\left(\mathrm{3}{x}\right)}{{x}}{dx} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \frac{\left({cosx}−{cos}\left(\mathrm{3}{x}\right)\right){e}^{−{ax}} }{{x}}{dx}\: \\ $$$$\left.{f}^{'} \left({a}\right)=−\int_{\mathrm{0}} ^{\infty} \left({cosx}\right){e}^{−{ax}} −{cos}\left(\mathrm{3}{x}\right){e}^{−{ax}} \right){dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} {cos}\left(\mathrm{3}{x}\right){e}^{−{ax}} {dx}−\int_{\mathrm{0}} ^{\infty} {cosx}\:{e}^{−{ax}} {dx} \\ $$$$={Re}\left(\int_{\mathrm{0}} ^{\infty} {e}^{\mathrm{3}{ix}−{ax}} {dx}\right)−{Re}\left(\int_{\mathrm{0}} ^{\infty} {e}^{{ix}−{ax}} {dx}\right) \\ $$$${but}\:\int_{\mathrm{0}} ^{\infty} {e}^{\left(−{a}+\mathrm{3}{i}\right){x}} {dx} \\ $$$$\left.=\frac{\mathrm{1}}{−{a}+\mathrm{3}{i}}{e}^{\left(−{a}+\mathrm{3}{i}\right)} \right]_{\mathrm{0}} ^{\infty} =\frac{\mathrm{1}}{{a}−\mathrm{3}{i}} \\ $$$$=\frac{{a}+\mathrm{3}{i}}{{a}^{\mathrm{2}} +\mathrm{9}}\:{and} \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{\left(−{a}+{i}\right){x}} {dx}=\left[\frac{\mathrm{1}}{−{a}+{i}}{e}^{\left(−{a}+{i}\right){x}} \right]_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{1}}{{a}−{i}}=\frac{{a}+{i}}{{a}^{\mathrm{2}} +\mathrm{1}}\:\Rightarrow{f}^{'} \left({a}\right)=\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{9}}−\frac{{a}}{{a}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow{f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{a}^{\mathrm{2}} +\mathrm{9}}{{a}^{\mathrm{2}} +\mathrm{1}}\right)+{k} \\ $$$${k}={lim}_{{a}\rightarrow\infty} {f}\left({a}\right)=\mathrm{0}\:\Rightarrow \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{a}^{\mathrm{2}} +\mathrm{9}}{{a}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{{cosx}−{cos}\left(\mathrm{3}{x}\right)}{{x}}{dx}={f}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{9} \\ $$$$={ln}\mathrm{3}\:\Rightarrow{I}=\frac{\mathrm{3}}{\mathrm{4}}{ln}\mathrm{3} \\ $$
Commented by mnjuly1970 last updated on 24/Jan/24
$$\:{thanks}\:{alot}\:{sir} \\ $$$$\:\:{so}\:{nice}\:{solution} \\ $$