Question Number 86374 by mathmax by abdo last updated on 28/Mar/20
$${calculate}\:{bycomplex}\:{method}\:\:\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$
Commented by mathmax by abdo last updated on 28/Mar/20
![∫_1 ^(+∞) (dx/(x^2 +1)) =∫_1 ^(+∞) (dx/((x−i)(x+i))) =(1/(2i))∫_1 ^(+∞) ((1/(x−i))−(1/(x+i)))dx =(1/(2i))[ln(((x−i)/(x+i)))]_1 ^(+∞) =(1/(2i))(−ln(((1−i)/(1+i)))) =(1/(2i))ln(((1+i)/(1−i))) we have ((1+i)/(1−i)) =(((√2)e^((iπ)/4) )/( (√2)e^(−((iπ)/4)) )) =e^((iπ)/2) = ⇒ln(((1+i)/(1−i))) =((iπ)/2) ⇒ ∫_1 ^(+∞) (dx/(x^2 +1)) =(1/(2i))×((iπ)/2) =(π/4)](https://www.tinkutara.com/question/Q86485.png)
$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:=\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{\left({x}−{i}\right)\left({x}+{i}\right)}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\int_{\mathrm{1}} ^{+\infty} \left(\frac{\mathrm{1}}{{x}−{i}}−\frac{\mathrm{1}}{{x}+{i}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left[{ln}\left(\frac{{x}−{i}}{{x}+{i}}\right)\right]_{\mathrm{1}} ^{+\infty} \:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(−{ln}\left(\frac{\mathrm{1}−{i}}{\mathrm{1}+{i}}\right)\right)\:=\frac{\mathrm{1}}{\mathrm{2}{i}}{ln}\left(\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}}\right)\:{we}\:{have} \\ $$$$\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}}\:=\frac{\sqrt{\mathrm{2}}{e}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:={e}^{\frac{{i}\pi}{\mathrm{2}}} \:=\:\Rightarrow{ln}\left(\frac{\mathrm{1}+{i}}{\mathrm{1}−{i}}\right)\:=\frac{{i}\pi}{\mathrm{2}}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{+\infty} \:\frac{{dx}}{{x}^{\mathrm{2}} \:+\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}{i}}×\frac{{i}\pi}{\mathrm{2}}\:=\frac{\pi}{\mathrm{4}} \\ $$