Integration Questions

Question Number 90192 by Rohit@Thakur last updated on 21/Apr/20

$$\int_{\mathrm{0}} ^{\mathrm{infinity}} \frac{\left(\mathrm{1}−\mathrm{e}^{−\mathrm{x}} \right)\mathrm{cosx}\:\mathrm{dx}}{\mathrm{x}} \\$$

Commented bymathmax by abdo last updated on 22/Apr/20

$${A}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{1}−{e}^{−{x}} \right){cosx}}{{x}}{dx}\:\Rightarrow{A}\:={Re}\left(\int_{\mathrm{0}} ^{\infty} \:\frac{\left(\mathrm{1}−{e}^{−{x}} \right){e}^{{ix}} }{{x}}{dx}\right) \\$$ $${we}\:{have}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{{ix}} −{e}^{\left(−\mathrm{1}+{i}\right){x}} }{{x}}{dx}\:={f}\left(−\mathrm{1}+{i}\right)\:{with} \\$$ $${f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{{ix}} −{e}^{{ax}} }{{x}}{dx}\:\:\:\left({a}\:{from}\:{C}\:\:{and}\:{Re}\left({a}\right)<\mathrm{0}\right) \\$$ $${f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:{e}^{{ax}} \:{dx}\:=−\frac{\mathrm{1}}{{a}}\left[{e}^{{ax}} \right]_{\mathrm{0}} ^{+\infty} =\frac{\mathrm{1}}{{a}}\:\Rightarrow{f}\left({a}\right)={ln}\left({a}\right)+{C} \\$$ $${lim}_{{a}\rightarrow{i}} {f}\left({a}\right)\:=\mathrm{0}\:={ln}\left({i}\right)+{C}\:={ln}\left({e}^{\frac{{i}\pi}{\mathrm{2}}} \right)\:+{C}\:=\frac{{i}\pi}{\mathrm{2}}\:+{C}\:\Rightarrow{C}=−\frac{{i}\pi}{\mathrm{2}} \\$$ $${f}\left({a}\right)\:={ln}\left({a}\right)−\frac{{i}\pi}{\mathrm{2}}\:\Rightarrow{f}\left(−\mathrm{1}+{i}\right)\:={ln}\left(−\mathrm{1}+{i}\right)−\frac{{i}\pi}{\mathrm{2}} \\$$ $$={ln}\left(\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)−\frac{{i}\pi}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−\frac{{i}\pi}{\mathrm{4}}−\frac{{i}\pi}{\mathrm{2}}\:=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}−\frac{\mathrm{3}{i}\pi}{\mathrm{4}} \\$$ $${A}\:={Re}\left(\int...\right)\:\Rightarrow{A}\:=\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}} \\$$

Answered by TANMAY PANACEA. last updated on 21/Apr/20

$${another}\:{approach}... \\$$ $$\int_{\mathrm{0}} ^{\infty} {e}^{−{ax}} {cosbx}\:{dx}=\frac{{a}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\$$ $$\int_{\mathrm{0}} ^{\infty} \left[\int_{\alpha} ^{\beta} {e}^{−{ax}} {da}\right]\:{cosbx}\:{dx}=\int_{\alpha} ^{\beta} \frac{{a}}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }{da} \\$$ $$\int_{\mathrm{0}} ^{\infty} \mid\frac{{e}^{−{ax}} }{−{x}}\mid_{\alpha} ^{\beta} {cosbx}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\alpha} ^{\beta} \frac{{d}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\$$ $$\int_{\mathrm{0}} ^{\infty} \left(\frac{{e}^{−\alpha{x}} −{e}^{−\beta{x}} }{{x}}\right){cosbx}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}\mid{ln}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)\mid_{\alpha} ^{\beta} \\$$ $$\int_{\mathrm{0}} ^{\infty} \left(\frac{{e}^{−\alpha{x}} −{e}^{−\beta{x}} }{{x}}\right){cosbx}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\beta^{\mathrm{2}} +{b}^{\mathrm{2}} }{\alpha^{\mathrm{2}} +{b}^{\mathrm{2}} }\right) \\$$ $${now}\:{put}\:\alpha=\mathrm{0}\:\:\:\beta=\mathrm{1}\:\:{b}=\mathrm{1} \\$$ $$\int_{\mathrm{0}} ^{\infty} \left(\frac{\mathrm{1}−{e}^{−{x}} }{{x}}\right){cosx}\:{dx}=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{\mathrm{1}+\mathrm{1}}{\mathrm{0}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\mathrm{2} \\$$ $${solved}\:{using}\:{ML}\:{khanna}\:{book}\:{formula} \\$$