Question Number 226799 by CrispyXYZ last updated on 15/Dec/25
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+{x}}\:\mathrm{d}{x}\:=\:? \\ $$
Answered by breniam last updated on 17/Dec/25
![2∫_0 ^1 y∫_0 ^1 (1/(x+1))×(x^2 /(x^2 y^2 +1))dxdy= 2∫_0 ^1 (y/(y^2 +1))∫_0 ^1 [(1/(x+1))+((x−1)/(x^2 y^2 +1))]dxdy=2∫_0 ^1 (y/(y^2 +1))[ln∣x+1∣+(1/(2y^2 ))ln∣x^2 y^2 +1∣−(1/y)arctan(xy)]_(x→0) ^(x→1) dy= =2∫_0 ^1 (y/(y^2 +1))(ln(2)+(1/(2y^2 ))ln(y^2 +1)−(1/y)arctan(y))dy= 2ln(2)∫_0 ^1 (y/(y^2 +1))dy+∫_0 ^1 (1/(y(y^2 +1)))ln(y^2 +1)dy−2∫_0 ^1 ((arctan(y))/(y^2 +1))dy ∫_0 ^1 (y/(y^2 +1))dy=(1/2)ln(2) ∫_0 ^1 (1/(y(y^2 +1)))ln(y^2 +1)dy={y^− =y^2 }=(1/2)∫_0 ^1 ((ln(y+1))/(y(y+1)))dy= (1/2)∫_0 ^1 ((ln(y+1))/y)dy−(1/2)∫_0 ^1 ((ln(y+1))/(y+1))dy ∫_0 ^1 ((ln(y+1))/(y+1))dy=((ln^2 (2))/2) ∫_0 ^1 ((ln(y+1))/y)dy=^(McLaurinSeries) ∫_0 ^1 (1/y)Σ_(n=1) ^∞ (1/n)×(−1)^(n−1) y^n dy=^(Lebesgue dominated convergence theorem) Σ_(n=1) ^∞ (1/n)×(−1)^n ∫_0 ^1 y^(n−1) dy=Σ_(n=1) ^∞ (((−1)^(n−1) )/n^2 )=Σ_(n=1) ^∞ (1/((2n−1)^2 ))−Σ_(n=1) ^∞ (1/((2n)^2 ))= Σ_(n=1) ^∞ (1/n^2 )−Σ_(n=1) ^∞ (1/((2n)^2 ))−Σ_(n=1) ^∞ (1/((2n)^2 ))=Σ_(n=1) ^∞ (1/n^2 )−(1/2)Σ_(n=1) ^∞ (1/n^2 )= (1/2)Σ_(n=1) ^∞ (1/n^2 )=(1/2)ζ(2)=(π^2 /(12)) ∫_0 ^1 ((arctan(y))/(y^2 +1))dy=(π^2 /(32)) ln(2)∫_0 ^1 (y/(y^2 +1))dy+(1/2)∫_0 ^1 (1/(y(y^2 +1)))ln(y^2 +1)dy−∫_0 ^1 ((arctan(y))/(y^2 +1))dy= ln^2 (2)−((ln^2 (2))/4)+(π^2 /(24))−(π^2 /(16))=((3ln^2 (2))/4)−(π^2 /(48))](https://www.tinkutara.com/question/Q226831.png)
$$\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{y}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{{x}+\mathrm{1}}×\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{x}\mathrm{d}{y}= \\ $$$$\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{y}}{{y}^{\mathrm{2}} +\mathrm{1}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\left[\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{1}}\right]\mathrm{d}{x}\mathrm{d}{y}=\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{y}}{{y}^{\mathrm{2}} +\mathrm{1}}\left[\mathrm{ln}\mid{x}+\mathrm{1}\mid+\frac{\mathrm{1}}{\mathrm{2}{y}^{\mathrm{2}} }\mathrm{ln}\mid{x}^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{1}\mid−\frac{\mathrm{1}}{{y}}\mathrm{arctan}\left({xy}\right)\right]_{{x}\rightarrow\mathrm{0}} ^{{x}\rightarrow\mathrm{1}} \mathrm{d}{y}= \\ $$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{y}}{{y}^{\mathrm{2}} +\mathrm{1}}\left(\mathrm{ln}\left(\mathrm{2}\right)+\frac{\mathrm{1}}{\mathrm{2}{y}^{\mathrm{2}} }\mathrm{ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)−\frac{\mathrm{1}}{{y}}\mathrm{arctan}\left({y}\right)\right)\mathrm{d}{y}= \\ $$$$\mathrm{2ln}\left(\mathrm{2}\right)\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{y}}{{y}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{y}+\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{{y}\left({y}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{d}{y}−\mathrm{2}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{arctan}\left({y}\right)}{{y}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{y} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{y}}{{y}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{y}=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{2}\right) \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{{y}\left({y}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{d}{y}=\left\{\overset{−} {{y}}={y}^{\mathrm{2}} \right\}=\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\left({y}+\mathrm{1}\right)}{{y}\left({y}+\mathrm{1}\right)}\mathrm{d}{y}= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\left({y}+\mathrm{1}\right)}{{y}}\mathrm{d}{y}−\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\left({y}+\mathrm{1}\right)}{{y}+\mathrm{1}}\mathrm{d}{y} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\left({y}+\mathrm{1}\right)}{{y}+\mathrm{1}}\mathrm{d}{y}=\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{2}} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{ln}\left({y}+\mathrm{1}\right)}{{y}}\mathrm{d}{y}\overset{\mathrm{McLaurinSeries}} {=}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{{y}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}×\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {y}^{{n}} \mathrm{d}{y}\overset{\mathrm{Lebesgue}\:\mathrm{dominated}\:\mathrm{convergence}\:\mathrm{theorem}} {=} \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}}×\left(−\mathrm{1}\right)^{{n}} \underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}{y}^{{n}−\mathrm{1}} \mathrm{d}{y}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}−\mathrm{1}\right)^{\mathrm{2}} }−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}} }= \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}} }−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\mathrm{2}{n}\right)^{\mathrm{2}} }=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\zeta\left(\mathrm{2}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{arctan}\left({y}\right)}{{y}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{y}=\frac{\pi^{\mathrm{2}} }{\mathrm{32}} \\ $$$$\mathrm{ln}\left(\mathrm{2}\right)\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{y}}{{y}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{y}+\frac{\mathrm{1}}{\mathrm{2}}\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{1}}{{y}\left({y}^{\mathrm{2}} +\mathrm{1}\right)}\mathrm{ln}\left({y}^{\mathrm{2}} +\mathrm{1}\right)\mathrm{d}{y}−\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{\mathrm{arctan}\left({y}\right)}{{y}^{\mathrm{2}} +\mathrm{1}}\mathrm{d}{y}= \\ $$$$\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)−\frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{4}}+\frac{\pi^{\mathrm{2}} }{\mathrm{24}}−\frac{\pi^{\mathrm{2}} }{\mathrm{16}}=\frac{\mathrm{3ln}^{\mathrm{2}} \left(\mathrm{2}\right)}{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{48}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by Tawa11 last updated on 16/Dec/25
$$\mathrm{Not}\:\mathrm{correct}\:\mathrm{sir}. \\ $$
Commented by breniam last updated on 17/Dec/25
$$\mathrm{Therebly}\:\mathrm{sorry}.\:\mathrm{I}\:\mathrm{started}\:\mathrm{to}\:\mathrm{fix}\:\mathrm{my}\:\mathrm{solution}, \\ $$$$\mathrm{but}\:\mathrm{concluding}\:\mathrm{from}\:\mathrm{another}\:\mathrm{comment},\:\mathrm{still}\:\mathrm{something}\:\mathrm{wrong}. \\ $$$$\mathrm{Working}\:\mathrm{on}\:\mathrm{making}\:\mathrm{it}\:\mathrm{fine}. \\ $$
Answered by Tawa11 last updated on 16/Dec/25
