# Question and Answers Forum

Integration Questions

Question Number 26949 by Joel578 last updated on 31/Dec/17

$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{xy}}\:{dx}\:{dy} \\$$

Commented by abdo imad last updated on 31/Dec/17

$$\left.{I}=\:\int_{\mathrm{0}^{} } ^{\mathrm{1}} \left(\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{dx}}{\mathrm{1}+{xy}}\right){dy}=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{{y}}{ln}\left(\mathrm{1}+{xy}\right)\right]_{{x}=\mathrm{0}} ^{{x}=\mathrm{1}} \right){dy} \\$$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{{ln}\left(\mathrm{1}+{y}\right)}{{y}}{dy}\:\:\:\:{but}\:\:\partial{ln}\left(\mathrm{1}+{y}\right)/\partial{y}=\:\frac{\mathrm{1}}{\mathrm{1}+{y}}\:=\:\sum_{{n}=\mathrm{0}} ^{\propto} \:\left(−\mathrm{1}\right)^{{n}} {y}^{{n}} \:\:\:\:{with}\:/{y}/<\mathrm{1} \\$$$${and}\:\:{ln}\left(\mathrm{1}+{y}\right)\:=\:\Sigma\:_{{n}=\mathrm{0}} ^{\propto} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+\mathrm{1}}\:{y}^{{n}+\mathrm{1}} \:=\:\sum_{{n}=\mathrm{1}} ^{\propto} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{y}^{{n}} \\$$$$\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{y}\right)}{{y}}{dy}\:=\:?\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{1}} ^{\propto} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{y}^{{n}−\mathrm{1}} \:\right) \\$$$$=\:\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:{but}\:\: \\$$$$\sum_{{n}=\mathrm{1}} ^{\propto} \:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:=\:−\sum_{{n}=\mathrm{1}} ^{\alpha} \:\frac{\mathrm{1}}{\mathrm{4}{n}^{\mathrm{2}} }\:+\sum_{{n}=\mathrm{0}} ^{\propto} \frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} } \\$$$$=\:−\frac{\mathrm{1}}{\mathrm{4}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:=\:\:\frac{\pi^{\mathrm{2}} }{\mathrm{12}}. \\$$

Commented by Joel578 last updated on 01/Jan/18

$${thank}\:{you}\:{very}\:{much} \\$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com