Menu Close

1-find-f-x-0-x-ln-t-ln-1-t-dt-with-0-x-1-2-find-the-value-of-0-1-ln-t-ln-1-t-dt-

Tinku Tara 0 Comments
Pin




Question Number 43918 by maxmathsup by imad last updated on 17/Sep/18
1) find f(x) =∫_0 ^x ln(t)ln(1−t)dt with 0≤x≤1 2) find the value of ∫_0 ^1 ln(t)ln(1−t)dt .
1)findf(x)=∫0xln(t)ln(1−t)dtwith0⩽x⩽12)findthevalueof∫01ln(t)ln(1−t)dt.
Commented by maxmathsup by imad last updated on 19/Sep/18
1) we have for 0<t≤x<1 ln^′ (1−t) =−(1/(1−t)) =−Σ_(n=0) ^∞ t^n ⇒ ln(1−t) =−Σ_(n=0) ^∞ (1/(n+1))t^(n+1) =−Σ_(n=1) ^∞ (t^n /n) ⇒ f(x) =−∫_0 ^x ln(t)(Σ_(n=1) ^∞ (t^n /n))dt =−Σ_(n=1) ^∞ (1/n) ∫_0 ^x t^n ln(t) dt by parts A_n (x)= ∫_0 ^x t^n ln(t)dt =[(1/(n+1))t^(n+1) ln(t)]_0 ^x −∫_0 ^x (1/(n+1))t^(n+1) (dt/t) =(1/(n+1)) x^(n+1) ln(x) −(1/(n+1))∫_0 ^x t^n dt = (1/(n+1))x^(n+1) ln(x)−(1/((n+1)^2 )) x^(n+1) ⇒ f(x) =−Σ_(n=1) ^∞ (1/n){ (1/(n+1)) x^(n+1) ln(x)−(1/((n+1)^2 )) x^(n+1) } = −Σ_(n=1) ^∞ (1/(n(n+1))) x^(n+1) ln(x) +Σ_(n=1) ^∞ (1/(n(n+1)^2 )) x^(n+1) =H(x)−K(x) K(x) =ln(x) Σ_(n=1) ^∞ {(1/n)−(1/(n+1))}x^(n+1) =ln(x)Σ_(n=1) ^∞ (x^(n+1) /n) −ln(x)Σ_(n=1) ^∞ (x^(n+1) /(n+1)) =xln(x) Σ_(n=1) ^∞ (x^n /n) −ln(x) Σ_(n=2) ^∞ (x^n /n) =−xln(x)ln(1−x) −ln(x)(−ln∣1−x∣−x) =−xln(x)ln∣1−x∣+ln(x)(ln∣1−x∣ +x) =(1−x)ln(x)ln∣1−x∣ +xln(x) .also we have (d/dx)K(x) =Σ_(n=1) ^∞ (1/(n(n+1))) x^n =Σ_(n=1) ^∞ {(1/n)−(1/(n+1))}x^n =Σ_(n=1) ^∞ (x^n /n) −Σ_(n=1) ^∞ (x^n /(n+1)) =−ln∣1−x∣−Σ_(n=2) ^∞ (x^(n−1) /n) =−ln∣1−x∣ −(1/x) {Σ_(n=1) ^∞ (x^n /n) −x} =−ln∣1−x∣+(1/x)ln∣1−x) +1 =((1/x)−1)ln∣1−x∣ +1 =(((1−x)ln∣1−x∣)/x) +1 ⇒ K(x) = x + ∫ (((1−x)ln∣1−x∣)/x) dx +k ....be continued...
1)wehavefor0<t⩽x<1ln′(1−t)=−11−t=−∑n=0∞tn⇒ln(1−t)=−∑n=0∞1n+1tn+1=−∑n=1∞tnn⇒f(x)=−∫0xln(t)(∑n=1∞tnn)dt=−∑n=1∞1n∫0xtnln(t)dtbypartsAn(x)=∫0xtnln(t)dt=[1n+1tn+1ln(t)]0x−∫0x1n+1tn+1dtt=1n+1xn+1ln(x)−1n+1∫0xtndt=1n+1xn+1ln(x)−1(n+1)2xn+1⇒f(x)=−∑n=1∞1n{1n+1xn+1ln(x)−1(n+1)2xn+1}=−∑n=1∞1n(n+1)xn+1ln(x)+∑n=1∞1n(n+1)2xn+1=H(x)−K(x)K(x)=ln(x)∑n=1∞{1n−1n+1}xn+1=ln(x)∑n=1∞xn+1n−ln(x)∑n=1∞xn+1n+1=xln(x)∑n=1∞xnn−ln(x)∑n=2∞xnn=−xln(x)ln(1−x)−ln(x)(−ln∣1−x∣−x)=−xln(x)ln∣1−x∣+ln(x)(ln∣1−x∣+x)=(1−x)ln(x)ln∣1−x∣+xln(x).alsowehaveddxK(x)=∑n=1∞1n(n+1)xn=∑n=1∞{1n−1n+1}xn=∑n=1∞xnn−∑n=1∞xnn+1=−ln∣1−x∣−∑n=2∞xn−1n=−ln∣1−x∣−1x{∑n=1∞xnn−x}=−ln∣1−x∣+1xln∣1−x)+1=(1x−1)ln∣1−x∣+1=(1−x)ln∣1−x∣x+1⇒K(x)=x+∫(1−x)ln∣1−x∣xdx+k….becontinued…
Commented by maxmathsup by imad last updated on 19/Sep/18
2) ∫_0 ^1 ln(t)ln(1−t)dt =lim_(x→0) ∫_0 ^x ln(t)ln(1−t)dt =lim_(x→1) { Σ_(n=1) ^∞ (x^(n+1) /((n+1)^2 )) −Σ_(n=1) ^∞ ((x^(n+1) lnx)/(n(n+1)))} =Σ_(n=1) ^∞ (1/((n+1)^2 )) =Σ_(n=2) ^∞ (1/n^2 ) =(π^2 /6) −1 .
2)∫01ln(t)ln(1−t)dt=limx→0∫0xln(t)ln(1−t)dt=limx→1{∑n=1∞xn+1(n+1)2−∑n=1∞xn+1lnxn(n+1)}=∑n=1∞1(n+1)2=∑n=2∞1n2=π26−1.
Commented by maxmathsup by imad last updated on 19/Sep/18
∫_0 ^1 ln(t)ln(1−t) dt =lim_(x→1) ∫_0 ^x ln(t)ln(1−t)dt =...
∫01ln(t)ln(1−t)dt=limx→1∫0xln(t)ln(1−t)dt=…

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *