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Question Number 201832 by sonukgindia last updated on 13/Dec/23

Answered by aleks041103 last updated on 13/Dec/23

$$x=e^{-t}\Rightarrow dx=-e^{-t}dt$$$$\Rightarrow I=32{\int }_0^{\mathrm{\infty }}\frac{{sin}^5(-t)}{-t}{e}^{-t}dt=32{\int }_0^{\mathrm{\infty }}\frac{e^{-t}{sin}^{5}t}{t}dt$$$$J\left(s\right)=-32{\int }_0^{\mathrm{\infty }}\frac{e^{-st}{sin}^{5}t}{t}dt,I=J\left(1\right)$$$$J{}^{\prime }\left(s\right)={\int }_0^{\mathrm{\infty }}-32{sin}^{5}t{e}^{-st}dt$$$$sin\left(t\right)=\frac{1}{2i}(e^{it}-e^{-it})$$$$-32{sin}^{5}t=-\frac{32}{2^{5}i}{(e^{it}-e^{-it})}^5=i{(e^{it}-e^{-it})}^5$$$$J{}^{\prime }=i{\int }_0^{\mathrm{\infty }}{e}^{-st}{(e^{it}-e^{-it})}^{5}dt=$$$$=i{\int }_0^{\mathrm{\infty }}{e}^{-(s+5i)t}{(e^{2it}-1)}^{5}dt=$$$$=i\underset{k=0}{\sum ^5}\left(\begin{array}{c}5\\ k\end{array}\right){(-1)}^{5-k}{\int }_0^{\mathrm{\infty }}{e}^{-(s+5i)t}{e}^{2ikt}dt$$$${\int }_0^{\mathrm{\infty }}{e}^{-(s+5i)t}{e}^{2ikt}dt={\int }_0^{\mathrm{\infty }}{e}^{(-s+(2k-5)i)t}dt=$$$$=\frac{1}{-s+(2k-5)i}{\left[e^{-st}{e}^{(sk-5)it}\right]}_0^{\mathrm{\infty }}=$$$$=\frac{-1}{(2k-5)i-s}$$$$\Rightarrow J{}^{\prime }=i\underset{k=0}{\sum ^5}\left(\begin{array}{c}5\\ k\end{array}\right){(-1)}^k\frac{1}{(2k-5)i-s}=$$$$=i(1.1.\frac{1}{-s-5i}+(-1).5.\frac{1}{-s-3i}+1.10.\frac{1}{-s-i}+(-1).10.\frac{1}{-s+i}+1.5.\frac{1}{-s+3i}-1.1.\frac{1}{-s+5i})=$$$$=i(-\frac{1}{s+5i}+\frac{5}{s+3i}-\frac{10}{s+i}+\frac{10}{s-i}-\frac{5}{s-3i}+\frac{1}{s-5i})=$$$$=i(\frac{1}{s-5i}-\frac{1}{s+5i})+5i(\frac{1}{s+3i}-\frac{1}{s-3i})+10i(\frac{1}{s-i}-\frac{1}{s+i})=$$$$=i\frac{10i}{s^2+25}+5i\frac{-6i}{s^2+9}+10i\frac{2i}{s^2+1}=$$$$=\frac{30}{s^2+9}-\frac{10}{s^2+25}-\frac{20}{s^2+1}$$$$J\left(\mathrm{\infty }\right)=0$$$$\Rightarrow I=J\left(1\right)-J\left(\mathrm{\infty }\right)={\int }_{\mathrm{\infty }}^{1}J{}^{\prime }\left(s\right)ds$$$$\Rightarrow I={\int }_1^{\mathrm{\infty }}(\frac{10}{s^2+5^2}+\frac{20}{s^2+1}-\frac{30}{s^2+3^2})ds$$$${\int }_1^{\mathrm{\infty }}\frac{ds}{s^2+a^2}=\frac{1}{a}{\int }_1^{\mathrm{\infty }}\frac{d\left(s/a\right)}{{\left(s/a\right)}^2+1}=$$$$=\frac{1}{a}[arctan\left(\mathrm{\infty }\right)-arctan\left(\frac{1}{a}\right)]=\frac{1}{a}(\frac{\pi }{2}-arctan\left(\frac{1}{a}\right))=$$$$=\frac{1}{a}arccot\left(\frac{1}{a}\right)=\frac{arctan\left(a\right)}{a}$$$$\Rightarrow I=\frac{10arctan\left(5\right)}{5}+\frac{20arctan\left(1\right)}{1}-\frac{30arctan\left(3\right)}{3}=$$$$=2arctan\left(5\right)+20\frac{\pi }{4}-10arctan\left(3\right)$$$$\Rightarrow {\int }_0^1\frac{32{sin}^5\left(lnx\right)}{lnx}dx=5\pi +2arctan\left(5\right)-10arctan\left(3\right)$$

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