Question Number 83852 by M±th+et£s last updated on 06/Mar/20
$${f}\left(\alpha\right)=\int_{\mathrm{0}} ^{\infty} \frac{{e}^{−\alpha{x}} {sin}\left({x}\right)}{{x}}{dx} \\ $$
Commented by mathmax by abdo last updated on 06/Mar/20
![for α≥0 we have f^′ (α) =−∫_0 ^∞ e^(−αx) sinx dx =−Im(∫_0 ^∞ e^(−αx+ix) dx) =−Im(∫_0 ^∞ e^((−α+i)x) dx) but ∫_0 ^∞ e^((−α+i)x) dx =[(1/(−α +i)) e^((−α+i)x) ]_0 ^(+∞) =(1/(α−i)) =((α+i)/(α^2 +1)) ⇒ f′(α) =−(1/(1+α^2 )) ⇒f(α)=−arctan(α)+K f(0) =∫_0 ^∞ ((sinx)/x)dx =(π/2) =K ⇒f(α)=(π/2) −arctan(α) (α≥0)](Q83856.png)
$${for}\:\alpha\geqslant\mathrm{0}\:\:{we}\:{have}\:{f}^{'} \left(\alpha\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\alpha{x}} \:{sinx}\:{dx} \\ $$$$=−{Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{−\alpha{x}+{ix}} {dx}\right)\:=−{Im}\left(\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−\alpha+{i}\right){x}} {dx}\right)\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{\left(−\alpha+{i}\right){x}} \:{dx}\:=\left[\frac{\mathrm{1}}{−\alpha\:+{i}}\:{e}^{\left(−\alpha+{i}\right){x}} \right]_{\mathrm{0}} ^{+\infty} \:=\frac{\mathrm{1}}{\alpha−{i}}\:=\frac{\alpha+{i}}{\alpha^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$${f}'\left(\alpha\right)\:=−\frac{\mathrm{1}}{\mathrm{1}+\alpha^{\mathrm{2}} }\:\Rightarrow{f}\left(\alpha\right)=−{arctan}\left(\alpha\right)+{K} \\ $$$${f}\left(\mathrm{0}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{sinx}}{{x}}{dx}\:=\frac{\pi}{\mathrm{2}}\:={K}\:\Rightarrow{f}\left(\alpha\right)=\frac{\pi}{\mathrm{2}}\:−{arctan}\left(\alpha\right)\:\:\:\:\left(\alpha\geqslant\mathrm{0}\right) \\ $$
Commented by M±th+et£s last updated on 07/Mar/20
$${thank}\:{you}\:{sir} \\ $$
Commented by abdomathmax last updated on 07/Mar/20
$${you}\:{are}\:{welcome}\:{sir} \\ $$