Question Number 31858 by NECx last updated on 16/Mar/18
$$\int\frac{{sinx}}{{x}}{dx} \\ $$
Commented by abdo imad last updated on 18/Mar/18
$${we}\:{have}\:{sinx}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:}{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:\:{with}\:{radius}\:{R}=\infty \\ $$$$\Rightarrow\:\frac{{sinx}}{{x}}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}\:{x}^{\mathrm{2}{n}} \:{and} \\ $$$$\int\frac{{sinx}}{{x}}{dx}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} \:\left(\mathrm{2}{n}\right)!}\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:\:{and}\:{this}\:{serie} \\ $$$${is}\:{convergent}\:…. \\ $$