Integration Questions

Question Number 97369 by  M±th+et+s last updated on 07/Jun/20

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left({ln}\left({x}\right)\right)^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\$$

Answered by maths mind last updated on 07/Jun/20

$${hello}\:{happy}\:{to}\:{comback}\: \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz}=\int_{\mathrm{1}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz} \\$$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz} \\$$$$\int_{−\infty} ^{+\infty} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz}=\mathrm{2}{i}\pi{Res}\left(\frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} },{z}={i}\right) \\$$$$\Rightarrow\int_{\mathrm{0}} ^{+\infty} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz}+\int_{−\infty} ^{\mathrm{0}} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz} \\$$$$=\int_{\mathrm{0}} ^{+\infty} \frac{{ln}^{\mathrm{2}} \left({z}\right)+{ln}^{\mathrm{2}} \left({z}\right)−\pi^{\mathrm{2}} +\mathrm{2}{i}\pi{ln}\left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz}=\mathrm{2}{i}\pi.\frac{{ln}^{\mathrm{2}} \left({i}\right)}{\mathrm{2}{i}}=\pi.−\frac{\pi^{\mathrm{2}} }{\mathrm{4}} \\$$$$\Leftrightarrow\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz}−\pi^{\mathrm{2}} \int_{\mathrm{0}} ^{+\infty} \frac{{dz}}{\mathrm{1}+{z}^{\mathrm{2}} }=−\frac{\pi^{\mathrm{3}} }{\mathrm{4}} \\$$$$\Rightarrow\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz}−\frac{\pi^{\mathrm{3}} }{\mathrm{2}}=−\frac{\pi^{\mathrm{3}} }{\mathrm{4}}\Leftrightarrow\int_{\mathrm{0}} ^{+\infty} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz}=\frac{\pi^{\mathrm{3}} }{\mathrm{8}} \\$$$$\Rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}^{\mathrm{2}} \left({z}\right)}{\mathrm{1}+{z}^{\mathrm{2}} }{dz}=\frac{\pi^{\mathrm{3}} }{\mathrm{16}} \\$$

Commented by  M±th+et+s last updated on 07/Jun/20

$${how}\:{are}\:{you}\:{sir}\:{wellllllcomeeeee}\:{back} \\$$$${very}\:{happy}\:{to}\:{see}\:{you}\:{again} \\$$

Commented by maths mind last updated on 07/Jun/20

$${im}\:{fin}\:{thanx}\:{i}\:{hop}\:{you}\:{too} \\$$$$\\$$$$\\$$