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Question-35551




Question Number 35551 by tanmay.chaudhury50@gmail.com last updated on 20/May/18
Commented by abdo mathsup 649 cc last updated on 21/May/18
let find the vslue of ∫_0 ^∞    ((e^(−ax)  −e^(−bx) )/x)dx with  a>0,b>0 let consider  f(t) =∫_0 ^∞  ((e^(−ax)  −e^(−bx) )/x)e^(−tx) d3  witht>0 after verifying the condition of  derivsbility we have   f^′ (t) = −∫_0 ^∞   e^(−tx) { e^(−ax)  −e^(−bx) }dx  =∫_0 ^∞   {e^(−(b+t)x)   −e^(−(a+t)x) }dx   =[((−1)/(b+t)) e^(−(b+t)x)   + (1/(a+t)) e^(−(a+t)x) ]_0 ^(+∞)   = (1/(b+t)) −(1/(a+t))  ⇒ f(t) = ln∣((b+t)/(a+t))∣ +c  but ∃m>0 /  ∣f(t)∣ ≤ m ∫_0 ^∞  e^(−tx) dx =(m/t) →0(t→+∞)   c =lim_(t→+∞) { f(t) −ln∣((b+t)/(a+t))∣} =0⇒  f(t) =ln (((b+t)/(a+t)))  and ∫_0 ^∞     ((e^(−ax)  −e^(−bx) )/x)dx  =f(0) =ln((b/a)) .
$${let}\:{find}\:{the}\:{vslue}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{ax}} \:−{e}^{−{bx}} }{{x}}{dx}\:{with} \\ $$$${a}>\mathrm{0},{b}>\mathrm{0}\:{let}\:{consider}\:\:{f}\left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{ax}} \:−{e}^{−{bx}} }{{x}}{e}^{−{tx}} {d}\mathrm{3} \\ $$$${witht}>\mathrm{0}\:{after}\:{verifying}\:{the}\:{condition}\:{of} \\ $$$${derivsbility}\:{we}\:{have}\: \\ $$$${f}^{'} \left({t}\right)\:=\:−\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−{tx}} \left\{\:{e}^{−{ax}} \:−{e}^{−{bx}} \right\}{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\left\{{e}^{−\left({b}+{t}\right){x}} \:\:−{e}^{−\left({a}+{t}\right){x}} \right\}{dx}\: \\ $$$$=\left[\frac{−\mathrm{1}}{{b}+{t}}\:{e}^{−\left({b}+{t}\right){x}} \:\:+\:\frac{\mathrm{1}}{{a}+{t}}\:{e}^{−\left({a}+{t}\right){x}} \right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\:\frac{\mathrm{1}}{{b}+{t}}\:−\frac{\mathrm{1}}{{a}+{t}}\:\:\Rightarrow\:{f}\left({t}\right)\:=\:{ln}\mid\frac{{b}+{t}}{{a}+{t}}\mid\:+{c}\:\:{but}\:\exists{m}>\mathrm{0}\:/ \\ $$$$\mid{f}\left({t}\right)\mid\:\leqslant\:{m}\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−{tx}} {dx}\:=\frac{{m}}{{t}}\:\rightarrow\mathrm{0}\left({t}\rightarrow+\infty\right)\: \\ $$$${c}\:={lim}_{{t}\rightarrow+\infty} \left\{\:{f}\left({t}\right)\:−{ln}\mid\frac{{b}+{t}}{{a}+{t}}\mid\right\}\:=\mathrm{0}\Rightarrow \\ $$$${f}\left({t}\right)\:={ln}\:\left(\frac{{b}+{t}}{{a}+{t}}\right)\:\:{and}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{e}^{−{ax}} \:−{e}^{−{bx}} }{{x}}{dx} \\ $$$$={f}\left(\mathrm{0}\right)\:={ln}\left(\frac{{b}}{{a}}\right)\:. \\ $$

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