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Question Number 38453 by maxmathsup by imad last updated on 25/Jun/18
find f(x)=∫_0 ^∞ ((1−cos(xt))/t) e^(−xt) dt with x>0 1) find asimple form of f(x) 2) calculate ∫_0 ^∞ ((1−cos(πt))/t) e^(−t) dt 3)calculate ∫_0 ^∞ ((1−cos(3t))/t) e^(−2t) dt
$${find}\:\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cos}\left({xt}\right)}{{t}}\:{e}^{−{xt}} {dt}\:{with}\:{x}>\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{asimple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left(\pi{t}\right)}{{t}}\:{e}^{−{t}} {dt} \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left(\mathrm{3}{t}\right)}{{t}}\:{e}^{−\mathrm{2}{t}} {dt} \\ $$
Commented by math khazana by abdo last updated on 26/Jun/18
the Q is find f(x)=∫_0 ^∞ ((1−cos(at))/t) e^(−xt) dt
$${the}\:{Q}\:{is}\:{find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cos}\left({at}\right)}{{t}}\:{e}^{−{xt}} {dt} \\ $$
Commented by math khazana by abdo last updated on 29/Jun/18
1) we have f(x)=∫_0 ^∞ ((1−cos(at))/t) e^(−xt) dt ⇒ f^′ (x)=−∫_0 ^∞ (1−cos(at))e^(−xt) dt =∫_0 ^∞ cos(at)e^(−xt) dt −∫_0 ^∞ e^(−xt) dt but ∫_0 ^∞ e^(−xt) dt =[−(1/x)e^(−xt) ]_0 ^(+∞) =(1/x) ∫_0 ^∞ cos(at) e^(−xt) dt =Re( ∫_0 ^∞ e^(iat−xt) dt) ∫_0 ^∞ e^((−x+ia)t) dt = [ (1/(−x+ia))e^((−x+ia)t) ]_0 ^(+∞) =((−1)/(−x+ia)) = (1/(x−ia)) = ((x+ia)/(x^2 +a^2 )) ⇒ ∫_0 ^∞ cos(at)e^(−xt) dt = (x/(x^2 +a^2 )) ⇒ f^′ (x)= (x/(x^2 +a^2 )) −(1/x) ⇒ f(x)= (1/2)ln(x^2 +a^2 ) −ln(x) +c but ∃m>0 < ∣ ∫_0 ^∞ ((1−cos(at))/t) e^(−xt) dt∣ ≤ m ∫_0 ^∞ e^(−xt) dt=(m/x) →0 when x→+∞ also we have f(x)=(1/2)ln(((x^2 +a^2 )/x^2 )) +c ⇒ c=lim_(x→+∞) f(x)−(1/2)ln(((x^2 +a^2 )/x^2 ) )=0 ⇒ f(x)=(1/2)ln(x^2 +a^2 ) −ln(x) with x>0
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left({at}\right)}{{t}}\:{e}^{−{xt}} {dt}\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=−\int_{\mathrm{0}} ^{\infty} \:\:\:\left(\mathrm{1}−{cos}\left({at}\right)\right){e}^{−{xt}} {dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:{cos}\left({at}\right){e}^{−{xt}} {dt}\:−\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{xt}} {dt}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} {dt}\:=\left[−\frac{\mathrm{1}}{{x}}{e}^{−{xt}} \right]_{\mathrm{0}} ^{+\infty} =\frac{\mathrm{1}}{{x}} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{cos}\left({at}\right)\:{e}^{−{xt}} {dt}\:={Re}\left(\:\int_{\mathrm{0}} ^{\infty} \:{e}^{{iat}−{xt}} {dt}\right) \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:{e}^{\left(−{x}+{ia}\right){t}} {dt}\:=\:\left[\:\frac{\mathrm{1}}{−{x}+{ia}}{e}^{\left(−{x}+{ia}\right){t}} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\frac{−\mathrm{1}}{−{x}+{ia}}\:=\:\frac{\mathrm{1}}{{x}−{ia}}\:=\:\frac{{x}+{ia}}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{cos}\left({at}\right){e}^{−{xt}} {dt}\:=\:\frac{{x}}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:\Rightarrow \\ $$$${f}^{'} \left({x}\right)=\:\frac{{x}}{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }\:−\frac{\mathrm{1}}{{x}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\:−{ln}\left({x}\right)\:+{c}\:\:{but}\:\exists{m}>\mathrm{0}\:< \\ $$$$\mid\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left({at}\right)}{{t}}\:{e}^{−{xt}} {dt}\mid\:\leqslant\:{m}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{xt}} {dt}=\frac{{m}}{{x}}\:\rightarrow\mathrm{0} \\ $$$${when}\:{x}\rightarrow+\infty\:{also}\:{we}\:{have}\: \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\right)\:+{c}\:\Rightarrow \\ $$$${c}={lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\frac{{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} }{{x}^{\mathrm{2}} }\:\right)=\mathrm{0}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)\:−{ln}\left({x}\right)\:\:{with}\:{x}>\mathrm{0} \\ $$
Commented by math khazana by abdo last updated on 29/Jun/18
2) we have proved that ∫_0 ^∞ ((1−cos(at))/t) e^(−xt) dt =(1/2)ln(x^2 +a^2 )−ln(x)⇒ ∫_0 ^∞ ((1−cos(πt))/t) e^(−t) =(1/2)ln(1+π^2 ).
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{proved}\:{that}\: \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left({at}\right)}{{t}}\:{e}^{−{xt}} {dt}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)−{ln}\left({x}\right)\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{cos}\left(\pi{t}\right)}{{t}}\:{e}^{−{t}} =\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{1}+\pi^{\mathrm{2}} \right). \\ $$
Commented by math khazana by abdo last updated on 29/Jun/18
3)?∫_0 ^∞ ((1−cos(3t))/t) e^(−2t) dt =(1/2)ln(2^2 +3^2 )−ln(2) =(1/2)ln(13)−ln(2).
$$\left.\mathrm{3}\right)?\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}−{cos}\left(\mathrm{3}{t}\right)}{{t}}\:{e}^{−\mathrm{2}{t}} {dt}\:=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{2}^{\mathrm{2}} \:+\mathrm{3}^{\mathrm{2}} \right)−{ln}\left(\mathrm{2}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left(\mathrm{13}\right)−{ln}\left(\mathrm{2}\right). \\ $$

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