Question Number 60595 by maxmathsup by imad last updated on 22/May/19
$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\left(\mathrm{1}−{ax}\right)^{\mathrm{2}} }\:{dx}\:\:{with}\:\mid{a}\mid<\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left({a}\right) \\ $$$$\left.\mathrm{2}\right)\:{determine}\:{A}\left(\theta\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{ln}^{\mathrm{2}} \left({x}\right)}{\left(\mathrm{1}−\left({cos}\theta\right){x}\right)^{\mathrm{2}} }{dx}\:\:{with}\:\mathrm{0}<\theta<\frac{\pi}{\mathrm{2}} \\ $$
Commented by maxmathsup by imad last updated on 23/May/19
![1) we have for ∣x∣<1 Σ_(n=0) ^∞ x^n =(1/(1−x)) and Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒ (1/((1−ax)^2 )) =Σ_(n=1) ^∞ n(ax)^(n−1) =Σ_(n=1) ^∞ n a^(n−1) x^(n−1) ⇒ f(a) =∫_0 ^1 (Σ_(n=1) ^∞ na^(n−1) x^(n−1) )ln^2 (x)dx =Σ_(n=1) ^∞ na^(n−1) ∫_0 ^1 x^(n−1) ln^2 (x)dx =Σ_(n=1) ^∞ na^(n−1) w_n with w_n =∫_0 ^1 x^(n−1) ln^2 (x)dx by parts u^′ =x^(n−1) and v=ln^2 x w_n =[(1/n)x^n ln^2 x]_0 ^1 −∫_0 ^1 (1/n)x^n ((2lnx)/x)dx =−(2/n) ∫_0 ^1 x^(n−1) ln(x) =_(byparts) −(2/n){ [(1/n)x^n lnx]_0 ^1 −∫_0 ^1 (1/n)x^n (dx/x)}=−(2/n){−(1/n) ∫_0 ^1 x^(n−1) dx} =(2/n^3 ) ⇒f(a) =Σ_(n=1) ^∞ ((2na^(n−1) )/n^3 ) =2 Σ_(n=1) ^∞ (a^(n−1) /n^2 ) ⇒af(a) =2Σ_(n=1) ^∞ (a^n /n^2 ) let try to find s(x) =Σ_(n=1) ^∞ (x^n /n^2 ) if ∣x∣<1 ....](https://www.tinkutara.com/question/Q60642.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{for}\:\mid{x}\mid<\mathrm{1}\:\:\:\:\:\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:{and}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} {nx}^{{n}−\mathrm{1}} \:=\frac{\mathrm{1}}{\left(\mathrm{1}−{x}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\left(\mathrm{1}−{ax}\right)^{\mathrm{2}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\left({ax}\right)^{{n}−\mathrm{1}} \:=\sum_{{n}=\mathrm{1}} ^{\infty} {n}\:{a}^{{n}−\mathrm{1}} {x}^{{n}−\mathrm{1}} \:\Rightarrow \\ $$$${f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\mathrm{1}} \left(\sum_{{n}=\mathrm{1}} ^{\infty} {na}^{{n}−\mathrm{1}} {x}^{{n}−\mathrm{1}} \right){ln}^{\mathrm{2}} \left({x}\right){dx}\:=\sum_{{n}=\mathrm{1}} ^{\infty} {na}^{{n}−\mathrm{1}} \int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{1}} {ln}^{\mathrm{2}} \left({x}\right){dx} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\infty} \:{na}^{{n}−\mathrm{1}} {w}_{{n}} \:\:\:\:\:{with}\:\:{w}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{1}} {ln}^{\mathrm{2}} \left({x}\right){dx}\:\:{by}\:{parts}\:{u}^{'} ={x}^{{n}−\mathrm{1}} \:{and}\:{v}={ln}^{\mathrm{2}} {x} \\ $$$${w}_{{n}} =\left[\frac{\mathrm{1}}{{n}}{x}^{{n}} {ln}^{\mathrm{2}} {x}\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{n}}{x}^{{n}} \:\frac{\mathrm{2}{lnx}}{{x}}{dx}\:=−\frac{\mathrm{2}}{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{1}} {ln}\left({x}\right)\:\: \\ $$$$=_{{byparts}} \:\:\:\:−\frac{\mathrm{2}}{{n}}\left\{\:\:\left[\frac{\mathrm{1}}{{n}}{x}^{{n}} {lnx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{n}}{x}^{{n}} \:\frac{{dx}}{{x}}\right\}=−\frac{\mathrm{2}}{{n}}\left\{−\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{n}−\mathrm{1}} {dx}\right\} \\ $$$$=\frac{\mathrm{2}}{{n}^{\mathrm{3}} }\:\Rightarrow{f}\left({a}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}{na}^{{n}−\mathrm{1}} }{{n}^{\mathrm{3}} }\:=\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{a}^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }\:\Rightarrow{af}\left({a}\right)\:=\mathrm{2}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{a}^{{n}} }{{n}^{\mathrm{2}} } \\ $$$${let}\:{try}\:{to}\:{find}\:\:{s}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}^{\mathrm{2}} }\:\:\:{if}\:\mid{x}\mid<\mathrm{1}\:\:\:…. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 23/May/19
![we have S^′ (x) =Σ_(n=1) ^∞ (x^(n−1) /n) ⇒x S^((1)) (x) =Σ_(n=1) ^∞ (x^n /n) ⇒ (x S^((1)) (x))^′ =Σ_(n=1) ^∞ x^(n−1) =Σ_(n=0) ^∞ x^n =(1/(1−x)) ⇒S^((1)) (x) +xS^((2)) (x) =(1/(1−x)) ⇒ S is solution of (de) xy^(′′) +y^′ =(1/(1−x)) let y^′ =z ⇒ xz^′ +z =(1/(1−x)) (e) (he) ⇒xz^′ +z =0 ⇒xz^′ =−z ⇒(z^′ /z) =−(1/x) ⇒ln∣z∣ =−ln∣x∣ +c ⇒ z =(k/(∣x∣)) let determine the solution on]0,+∞[ ⇒z =(k/x) ⇒ mvc method give z^′ =(k^′ /x) −(k/x^2 ) (e) ⇒k^′ −(k/x) +(k/x) =(1/(1−x)) ⇒k^′ =(1/(1−x)) ⇒k(x) =−ln(1−x)+c_0 ⇒z(x) =−((ln(1−x))/x) +(c_0 /x) y^′ =z ⇒y^′ =−((ln(1−x))/x) +(c_0 /x) ⇒ y(x) =−∫_0 ^x ((ln(1−t))/t) dt +c_0 ln(x) +λ ⇒ S(x) = c_0 ln(x)−∫_0 ^x ((ln(1−t))/t) dt ( x>0) S(e) =c_0 −∫_0 ^e ((ln(1−t))/t) dt =Σ_(n=1) ^∞ (e^n /n^2 ) ⇒c_0 =Σ_(n=1) ^∞ (e^n /n^2 ) +∫_0 ^e ((ln(1−t))/t) dt ⇒ S(x) =(Σ_(n=1) ^∞ (e^n /n^2 ) +∫_0 ^e ((ln(1−t))/t) dt)ln(x)−∫_0 ^x ((ln(1−t))/t) dt ....be continued...](https://www.tinkutara.com/question/Q60649.png)
$${we}\:{have}\:{S}^{'} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}−\mathrm{1}} }{{n}}\:\Rightarrow{x}\:{S}^{\left(\mathrm{1}\right)} \left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:\Rightarrow \\ $$$$\left({x}\:{S}^{\left(\mathrm{1}\right)} \left({x}\right)\right)^{'} \:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{x}^{{n}−\mathrm{1}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow{S}^{\left(\mathrm{1}\right)} \left({x}\right)\:+{xS}^{\left(\mathrm{2}\right)} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow \\ $$$${S}\:{is}\:{solution}\:{of}\:\left({de}\right)\:\:\:\:\:\:{xy}^{''} \:+{y}^{'} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:\:\:{let}\:{y}^{'} \:={z}\:\Rightarrow \\ $$$${xz}^{'} \:+{z}\:\:\:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\:\:\left({e}\right) \\ $$$$\left({he}\right)\:\Rightarrow{xz}^{'} \:+{z}\:=\mathrm{0}\:\Rightarrow{xz}^{'} \:=−{z}\:\Rightarrow\frac{{z}^{'} }{{z}}\:=−\frac{\mathrm{1}}{{x}}\:\Rightarrow{ln}\mid{z}\mid\:=−{ln}\mid{x}\mid\:+{c}\:\Rightarrow \\ $$$$\left.{z}\:=\frac{{k}}{\mid{x}\mid}\:\:\:\:{let}\:{determine}\:{the}\:{solution}\:{on}\right]\mathrm{0},+\infty\left[\:\Rightarrow{z}\:=\frac{{k}}{{x}}\:\Rightarrow\:{mvc}\:{method}\:{give}\right. \\ $$$${z}^{'} \:=\frac{{k}^{'} }{{x}}\:−\frac{{k}}{{x}^{\mathrm{2}} }\:\:\:\:\:\:\:\left({e}\right)\:\Rightarrow{k}^{'} \:−\frac{{k}}{{x}}\:+\frac{{k}}{{x}}\:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow{k}^{'} \:=\frac{\mathrm{1}}{\mathrm{1}−{x}}\:\Rightarrow{k}\left({x}\right)\:=−{ln}\left(\mathrm{1}−{x}\right)+{c}_{\mathrm{0}} \\ $$$$\Rightarrow{z}\left({x}\right)\:=−\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}\:+\frac{{c}_{\mathrm{0}} }{{x}} \\ $$$${y}^{'} ={z}\:\Rightarrow{y}^{'} \:=−\frac{{ln}\left(\mathrm{1}−{x}\right)}{{x}}\:+\frac{{c}_{\mathrm{0}} }{{x}}\:\Rightarrow\:{y}\left({x}\right)\:=−\int_{\mathrm{0}} ^{{x}} \:\:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:{dt}\:+{c}_{\mathrm{0}} {ln}\left({x}\right)\:+\lambda\:\Rightarrow \\ $$$${S}\left({x}\right)\:=\:{c}_{\mathrm{0}} {ln}\left({x}\right)−\int_{\mathrm{0}} ^{{x}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:{dt}\:\:\:\:\left(\:\:{x}>\mathrm{0}\right) \\ $$$${S}\left({e}\right)\:={c}_{\mathrm{0}} \:−\int_{\mathrm{0}} ^{{e}} \:\:\:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:{dt}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{n}} }{{n}^{\mathrm{2}} }\:\:\Rightarrow{c}_{\mathrm{0}} =\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{n}} }{{n}^{\mathrm{2}} }\:+\int_{\mathrm{0}} ^{{e}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:{dt}\:\Rightarrow \\ $$$${S}\left({x}\right)\:=\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{n}} }{{n}^{\mathrm{2}} }\:+\int_{\mathrm{0}} ^{{e}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:{dt}\right){ln}\left({x}\right)−\int_{\mathrm{0}} ^{{x}} \:\frac{{ln}\left(\mathrm{1}−{t}\right)}{{t}}\:{dt}\:….{be}\:{continued}… \\ $$