Question Number 33979 by abdo imad last updated on 28/Apr/18
$${we}\:{give}\:{for}\:{t}>\mathrm{0}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{sinx}}{{x}}\:{e}^{−{tx}} {dx}\:=\frac{\pi}{\mathrm{2}}\:−{arctant} \\ $$$${use}\:{this}\:{result}\:{to}\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{1}−{e}^{−{x}} \right){sinx}}{{x}^{\mathrm{2}} }{dx}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 03/May/18
![we know that ∫_0 ^∞ ((sinx)/x) e^(−tx) dx=(π/2) −arctant ⇒ ∫_0 ^1 ((π/2) −arctant)dt = ∫_0 ^1 ( ∫_0 ^∞ ((sinx)/x) e^(−tx) dx) =∫_0 ^∞ (∫_0 ^1 e^(−tx) dt) ((sinx)/x)dx ( by fubini) =∫_0 ^∞ ([−(1/x) e^(−tx) ]_(t=0) ^(t=1) )((sinx)/x)dx =∫_0 ^∞ (1/x^2 )(1−e^(−x) ) sinx dx but ∫_0 ^1 ( (π/2) −arctant)dt =(π/2) −∫_0 ^1 arctan t dt by parts ∫_0 ^1 arctan t dt = [tarctant]_0 ^1 −∫_0 ^1 (t/(1+t^2 ))dt =(π/4) −(1/2)[ln(1+t^2 )]_0 ^1 = (π/4) −(1/2)ln(2) ⇒ ∫_0 ^∞ (((1−e^(−x) )sinx)/x^2 ) dx = (π/4) +(1/2)ln(2) .](https://www.tinkutara.com/question/Q34235.png)
$${we}\:{know}\:{that}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sinx}}{{x}}\:{e}^{−{tx}} \:{dx}=\frac{\pi}{\mathrm{2}}\:−{arctant}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\frac{\pi}{\mathrm{2}}\:−{arctant}\right){dt}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\left(\:\int_{\mathrm{0}} ^{\infty} \:\frac{{sinx}}{{x}}\:{e}^{−{tx}} {dx}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\left(\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{e}^{−{tx}} {dt}\right)\:\frac{{sinx}}{{x}}{dx}\:\left(\:{by}\:{fubini}\right) \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\left(\left[−\frac{\mathrm{1}}{{x}}\:{e}^{−{tx}} \right]_{{t}=\mathrm{0}} ^{{t}=\mathrm{1}} \right)\frac{{sinx}}{{x}}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{{x}^{\mathrm{2}} }\left(\mathrm{1}−{e}^{−{x}} \right)\:{sinx}\:{dx}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left(\:\frac{\pi}{\mathrm{2}}\:−{arctant}\right){dt}\:=\frac{\pi}{\mathrm{2}}\:−\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{arctan}\:{t}\:{dt} \\ $$$${by}\:{parts}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:{arctan}\:{t}\:{dt}\:=\:\left[{tarctant}\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\overset{\mathrm{1}} {\int}_{\mathrm{0}} \:\frac{{t}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}\left[{ln}\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} \:\:=\:\frac{\pi}{\mathrm{4}}\:−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\left(\mathrm{1}−{e}^{−{x}} \right){sinx}}{{x}^{\mathrm{2}} }\:{dx}\:=\:\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\:. \\ $$