Question Number 201832 by sonukgindia last updated on 13/Dec/23
Answered by aleks041103 last updated on 13/Dec/23
![x=e^(−t) ⇒ dx=−e^(−t) dt ⇒I=32∫_0 ^( ∞) ((sin^5 (−t))/(−t))e^(−t) dt=32∫_0 ^( ∞) ((e^(−t) sin^5 t)/t)dt J(s)=−32∫_0 ^( ∞) ((e^(−st) sin^5 t)/t)dt, I=J(1) J ′(s)=∫_0 ^( ∞) −32sin^5 t e^(−st) dt sin(t)=(1/(2i))(e^(it) −e^(−it) ) −32sin^5 t=−((32)/(2^5 i))(e^(it) −e^(−it) )^5 =i(e^(it) −e^(−it) )^5 J ′=i∫_0 ^( ∞) e^(−st) (e^(it) −e^(−it) )^5 dt= =i∫_0 ^( ∞) e^(−(s+5i)t) (e^(2it) −1)^5 dt= =iΣ_(k=0) ^5 ((5),(k) ) (−1)^(5−k) ∫_0 ^( ∞) e^(−(s+5i)t) e^(2ikt) dt ∫_0 ^( ∞) e^(−(s+5i)t) e^(2ikt) dt=∫_0 ^( ∞) e^((−s+(2k−5)i)t) dt= =(1/(−s+(2k−5)i))[e^(−st) e^((sk−5)it) ]_0 ^∞ = =((−1)/((2k−5)i−s)) ⇒J ′=iΣ_(k=0) ^5 ((5),(k) ) (−1)^k (1/((2k−5)i−s))= =i(1.1.(1/(−s−5i))+(−1).5.(1/(−s−3i))+1.10.(1/(−s−i))+(−1).10.(1/(−s+i))+1.5.(1/(−s+3i))−1.1.(1/(−s+5i)))= =i(−(1/(s+5i))+(5/(s+3i))−((10)/(s+i))+((10)/(s−i))−(5/(s−3i))+(1/(s−5i)))= =i((1/(s−5i))−(1/(s+5i)))+5i((1/(s+3i))−(1/(s−3i)))+10i((1/(s−i))−(1/(s+i)))= =i((10i)/(s^2 +25))+5i((−6i)/(s^2 +9))+10i((2i)/(s^2 +1))= =((30)/(s^2 +9))−((10)/(s^2 +25))−((20)/(s^2 +1)) J(∞)=0 ⇒I=J(1)−J(∞)=∫_∞ ^( 1) J ′(s)ds ⇒I=∫_1 ^( ∞) (((10)/(s^2 +5^2 ))+((20)/(s^2 +1))−((30)/(s^2 +3^2 )))ds ∫_1 ^∞ (ds/(s^2 +a^2 ))=(1/a)∫_1 ^( ∞) ((d(s/a))/((s/a)^2 +1))= =(1/a)[arctan(∞)−arctan((1/a))]=(1/a)((π/2)−arctan((1/a)))= =(1/a)arccot((1/a))=((arctan(a))/a) ⇒I=((10arctan(5))/5)+((20arctan(1))/1)−((30arctan(3))/3)= =2arctan(5)+20(π/4)−10arctan(3) ⇒∫_0 ^( 1) ((32sin^5 (ln x))/(ln x))dx = 5π+2arctan(5)−10arctan(3)](https://www.tinkutara.com/question/Q201836.png)
$${x}={e}^{−{t}} \:\Rightarrow\:{dx}=−{e}^{−{t}} {dt} \\ $$$$\Rightarrow{I}=\mathrm{32}\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}^{\mathrm{5}} \left(−{t}\right)}{−{t}}{e}^{−{t}} {dt}=\mathrm{32}\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{−{t}} {sin}^{\mathrm{5}} {t}}{{t}}{dt} \\ $$$${J}\left({s}\right)=−\mathrm{32}\int_{\mathrm{0}} ^{\:\infty} \frac{{e}^{−{st}} {sin}^{\mathrm{5}} {t}}{{t}}{dt},\:{I}={J}\left(\mathrm{1}\right) \\ $$$${J}\:'\left({s}\right)=\int_{\mathrm{0}} ^{\:\infty} −\mathrm{32}{sin}^{\mathrm{5}} {t}\:{e}^{−{st}} {dt} \\ $$$${sin}\left({t}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}}\left({e}^{{it}} −{e}^{−{it}} \right) \\ $$$$−\mathrm{32}{sin}^{\mathrm{5}} {t}=−\frac{\mathrm{32}}{\mathrm{2}^{\mathrm{5}} {i}}\left({e}^{{it}} −{e}^{−{it}} \right)^{\mathrm{5}} ={i}\left({e}^{{it}} −{e}^{−{it}} \right)^{\mathrm{5}} \\ $$$${J}\:'={i}\int_{\mathrm{0}} ^{\:\infty} {e}^{−{st}} \left({e}^{{it}} −{e}^{−{it}} \right)^{\mathrm{5}} {dt}= \\ $$$$={i}\int_{\mathrm{0}} ^{\:\infty} {e}^{−\left({s}+\mathrm{5}{i}\right){t}} \left({e}^{\mathrm{2}{it}} −\mathrm{1}\right)^{\mathrm{5}} {dt}= \\ $$$$={i}\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\begin{pmatrix}{\mathrm{5}}\\{{k}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{\mathrm{5}−{k}} \int_{\mathrm{0}} ^{\:\infty} {e}^{−\left({s}+\mathrm{5}{i}\right){t}} {e}^{\mathrm{2}{ikt}} {dt} \\ $$$$\int_{\mathrm{0}} ^{\:\infty} {e}^{−\left({s}+\mathrm{5}{i}\right){t}} {e}^{\mathrm{2}{ikt}} {dt}=\int_{\mathrm{0}} ^{\:\infty} {e}^{\left(−{s}+\left(\mathrm{2}{k}−\mathrm{5}\right){i}\right){t}} {dt}= \\ $$$$=\frac{\mathrm{1}}{−{s}+\left(\mathrm{2}{k}−\mathrm{5}\right){i}}\left[{e}^{−{st}} {e}^{\left({sk}−\mathrm{5}\right){it}} \right]_{\mathrm{0}} ^{\infty} = \\ $$$$=\frac{−\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{5}\right){i}−{s}} \\ $$$$\Rightarrow{J}\:'={i}\underset{{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\begin{pmatrix}{\mathrm{5}}\\{{k}}\end{pmatrix}\:\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{5}\right){i}−{s}}= \\ $$$$={i}\left(\mathrm{1}.\mathrm{1}.\frac{\mathrm{1}}{−{s}−\mathrm{5}{i}}+\left(−\mathrm{1}\right).\mathrm{5}.\frac{\mathrm{1}}{−{s}−\mathrm{3}{i}}+\mathrm{1}.\mathrm{10}.\frac{\mathrm{1}}{−{s}−{i}}+\left(−\mathrm{1}\right).\mathrm{10}.\frac{\mathrm{1}}{−{s}+{i}}+\mathrm{1}.\mathrm{5}.\frac{\mathrm{1}}{−{s}+\mathrm{3}{i}}−\mathrm{1}.\mathrm{1}.\frac{\mathrm{1}}{−{s}+\mathrm{5}{i}}\right)= \\ $$$$={i}\left(−\frac{\mathrm{1}}{{s}+\mathrm{5}{i}}+\frac{\mathrm{5}}{{s}+\mathrm{3}{i}}−\frac{\mathrm{10}}{{s}+{i}}+\frac{\mathrm{10}}{{s}−{i}}−\frac{\mathrm{5}}{{s}−\mathrm{3}{i}}+\frac{\mathrm{1}}{{s}−\mathrm{5}{i}}\right)= \\ $$$$={i}\left(\frac{\mathrm{1}}{{s}−\mathrm{5}{i}}−\frac{\mathrm{1}}{{s}+\mathrm{5}{i}}\right)+\mathrm{5}{i}\left(\frac{\mathrm{1}}{{s}+\mathrm{3}{i}}−\frac{\mathrm{1}}{{s}−\mathrm{3}{i}}\right)+\mathrm{10}{i}\left(\frac{\mathrm{1}}{{s}−{i}}−\frac{\mathrm{1}}{{s}+{i}}\right)= \\ $$$$={i}\frac{\mathrm{10}{i}}{{s}^{\mathrm{2}} +\mathrm{25}}+\mathrm{5}{i}\frac{−\mathrm{6}{i}}{{s}^{\mathrm{2}} +\mathrm{9}}+\mathrm{10}{i}\frac{\mathrm{2}{i}}{{s}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{30}}{{s}^{\mathrm{2}} +\mathrm{9}}−\frac{\mathrm{10}}{{s}^{\mathrm{2}} +\mathrm{25}}−\frac{\mathrm{20}}{{s}^{\mathrm{2}} +\mathrm{1}} \\ $$$${J}\left(\infty\right)=\mathrm{0} \\ $$$$\Rightarrow{I}={J}\left(\mathrm{1}\right)−{J}\left(\infty\right)=\int_{\infty} ^{\:\mathrm{1}} {J}\:'\left({s}\right){ds} \\ $$$$\Rightarrow{I}=\int_{\mathrm{1}} ^{\:\infty} \left(\frac{\mathrm{10}}{{s}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }+\frac{\mathrm{20}}{{s}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{30}}{{s}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }\right){ds} \\ $$$$\int_{\mathrm{1}} ^{\infty} \frac{{ds}}{{s}^{\mathrm{2}} +{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{{a}}\int_{\mathrm{1}} ^{\:\infty} \frac{{d}\left({s}/{a}\right)}{\left({s}/{a}\right)^{\mathrm{2}} +\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{{a}}\left[{arctan}\left(\infty\right)−{arctan}\left(\frac{\mathrm{1}}{{a}}\right)\right]=\frac{\mathrm{1}}{{a}}\left(\frac{\pi}{\mathrm{2}}−{arctan}\left(\frac{\mathrm{1}}{{a}}\right)\right)= \\ $$$$=\frac{\mathrm{1}}{{a}}{arccot}\left(\frac{\mathrm{1}}{{a}}\right)=\frac{{arctan}\left({a}\right)}{{a}} \\ $$$$\Rightarrow{I}=\frac{\mathrm{10}{arctan}\left(\mathrm{5}\right)}{\mathrm{5}}+\frac{\mathrm{20}{arctan}\left(\mathrm{1}\right)}{\mathrm{1}}−\frac{\mathrm{30}{arctan}\left(\mathrm{3}\right)}{\mathrm{3}}= \\ $$$$=\mathrm{2}{arctan}\left(\mathrm{5}\right)+\mathrm{20}\frac{\pi}{\mathrm{4}}−\mathrm{10}{arctan}\left(\mathrm{3}\right) \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{32}{sin}^{\mathrm{5}} \left({ln}\:{x}\right)}{{ln}\:{x}}{dx}\:=\:\mathrm{5}\pi+\mathrm{2}{arctan}\left(\mathrm{5}\right)−\mathrm{10}{arctan}\left(\mathrm{3}\right) \\ $$