Question Number 67462 by ~ À ® @ 237 ~ last updated on 27/Aug/19
$$ \\ $$$$\:{Calculate}\:{when}\:{a},{b}\:{are}\:{positive}\:{reals}\:\:\:{f}\left({a},{b}\right)=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{t}^{{a}} −{t}^{{b}} }{{lnt}}\:{dt}\: \\ $$
Commented by mathmax by abdo last updated on 27/Aug/19
![changement lnt =−x give f(a,b) =−∫_0 ^(+∞) ((e^(−ax) −e^(−bx) )/(−x))(−e^(−x) )dx =−∫_0 ^∞ ((e^(−(a+1)x) −e^(−(b+1)x) )/x)dx =∫_0 ^∞ ((e^(−(b+1)x) −e^(−(a+1)x) )/x)dx we have (∂f/∂a)(a,b) =∫_0 ^∞ ((xe^(−(a+1)x) )/x)dx =∫_0 ^∞ e^(−(a+1)x) dx =[((−1)/(a+1)) e^(−(a+1)x) ]_0 ^(+∞) =−(1/(a+1)){−1} =(1/(a+1)) ⇒ f(a,b) =ln(a+1) +c we have f(0,b) =c =∫_0 ^1 ((1−t^b )/(lnt))dt =_(lnt =−u) −∫_0 ^∞ ((1−e^(−bt) )/(−u))(−e^(−u) )du =−∫_0 ^∞ ((e^(−u) −e^(−(b+1)u) )/u)du =∫_0 ^∞ ((e^(−(b+1)u) −e^(−u) )/u)du =ϕ(b) we have ϕ^′ (b) =−∫_0 ^∞ e^(−(b+1)u) du =[(1/(b+1)) e^(−(b+1)u) ]_0 ^(+∞) =−(1/(b+1)) ⇒ ϕ(b) =−ln(b+1) +c with c=ϕ(0) =0 ⇒ f(0,b) =−ln(b+1) ⇒f(a,b) =ln(a+1)−ln(b+1) =ln(((a+1)/(b+1))).](Q67483.png)
$${changement}\:{lnt}\:=−{x}\:{give}\:{f}\left({a},{b}\right)\:=−\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{e}^{−{ax}} −{e}^{−{bx}} }{−{x}}\left(−{e}^{−{x}} \right){dx} \\ $$$$=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\left({a}+\mathrm{1}\right){x}} −{e}^{−\left({b}+\mathrm{1}\right){x}} }{{x}}{dx}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{e}^{−\left({b}+\mathrm{1}\right){x}} −{e}^{−\left({a}+\mathrm{1}\right){x}} }{{x}}{dx} \\ $$$${we}\:{have}\:\frac{\partial{f}}{\partial{a}}\left({a},{b}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{{xe}^{−\left({a}+\mathrm{1}\right){x}} }{{x}}{dx}\:=\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({a}+\mathrm{1}\right){x}} {dx} \\ $$$$=\left[\frac{−\mathrm{1}}{{a}+\mathrm{1}}\:{e}^{−\left({a}+\mathrm{1}\right){x}} \right]_{\mathrm{0}} ^{+\infty} \:=−\frac{\mathrm{1}}{{a}+\mathrm{1}}\left\{−\mathrm{1}\right\}\:=\frac{\mathrm{1}}{{a}+\mathrm{1}}\:\Rightarrow \\ $$$${f}\left({a},{b}\right)\:={ln}\left({a}+\mathrm{1}\right)\:+{c}\:\:\:{we}\:{have}\:{f}\left(\mathrm{0},{b}\right)\:={c}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}−{t}^{{b}} }{{lnt}}{dt} \\ $$$$=_{{lnt}\:=−{u}} \:\:\:\:−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}−{e}^{−{bt}} }{−{u}}\left(−{e}^{−{u}} \right){du}\:=−\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−{u}} −{e}^{−\left({b}+\mathrm{1}\right){u}} }{{u}}{du} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{e}^{−\left({b}+\mathrm{1}\right){u}} −{e}^{−{u}} }{{u}}{du}\:=\varphi\left({b}\right)\:\:{we}\:{have}\: \\ $$$$\varphi^{'} \left({b}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({b}+\mathrm{1}\right){u}} {du}\:=\left[\frac{\mathrm{1}}{{b}+\mathrm{1}}\:{e}^{−\left({b}+\mathrm{1}\right){u}} \right]_{\mathrm{0}} ^{+\infty} \:=−\frac{\mathrm{1}}{{b}+\mathrm{1}}\:\Rightarrow \\ $$$$\varphi\left({b}\right)\:=−{ln}\left({b}+\mathrm{1}\right)\:+{c}\:\:\:\:\:{with}\:\:{c}=\varphi\left(\mathrm{0}\right)\:=\mathrm{0}\:\Rightarrow \\ $$$${f}\left(\mathrm{0},{b}\right)\:=−{ln}\left({b}+\mathrm{1}\right)\:\Rightarrow{f}\left({a},{b}\right)\:={ln}\left({a}+\mathrm{1}\right)−{ln}\left({b}+\mathrm{1}\right) \\ $$$$={ln}\left(\frac{{a}+\mathrm{1}}{{b}+\mathrm{1}}\right). \\ $$
Commented by ~ À ® @ 237 ~ last updated on 28/Aug/19
$${Thanks}\:{you}\:{sir} \\ $$