Question Number 200549 by sonukgindia last updated on 20/Nov/23
Answered by witcher3 last updated on 20/Nov/23
![I_(12) =0;x→(1/x) I_(11) =∫_0 ^1 ((ln^2 (x))/(1+x^2 ))+∫_1 ^∞ ((ln^2 (x))/(1+x^2 ))dx x→(1/x) =2∫_0 ^1 ((ln^2 (z))/(1+z^2 ))dz=2∫_0 ^1 ln^2 (z)(Σ_(n≥1) (−1)^n z^(2n) libneiz theorem Σ(−1)^n z^(2n) cv normaly in [0,a] a<1 ⇒I_(12) =2Σ_(n≥0) (−1)^n ∫_0 ^1 z^(2n) ln^2 (z) =4Σ_(n≥1) (((−1)^n )/((2n+1)^3 ))= =4(π^3 /(32))=(π^3 /8)](Q200571.png)
$$\mathrm{I}_{\mathrm{12}} =\mathrm{0};\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{I}_{\mathrm{11}} =\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\mathrm{x}\rightarrow\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{ln}^{\mathrm{2}} \left(\mathrm{z}\right)}{\mathrm{1}+\mathrm{z}^{\mathrm{2}} }\mathrm{dz}=\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{z}\right)\left(\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{z}^{\mathrm{2n}} \right. \\ $$$$\mathrm{libneiz}\:\mathrm{theorem}\:\Sigma\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{z}^{\mathrm{2n}} \:\mathrm{cv}\:\mathrm{normaly}\:\mathrm{in}\:\left[\mathrm{0},\mathrm{a}\right] \\ $$$$\mathrm{a}<\mathrm{1} \\ $$$$\Rightarrow\mathrm{I}_{\mathrm{12}} =\mathrm{2}\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{n}} \int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{z}^{\mathrm{2n}} \mathrm{ln}^{\mathrm{2}} \left(\mathrm{z}\right) \\ $$$$=\mathrm{4}\underset{\mathrm{n}\geqslant\mathrm{1}} {\sum}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{1}\right)^{\mathrm{3}} }= \\ $$$$=\mathrm{4}\frac{\pi^{\mathrm{3}} }{\mathrm{32}}=\frac{\pi^{\mathrm{3}} }{\mathrm{8}} \\ $$
Commented by sonukgindia last updated on 20/Nov/23
$${nice} \\ $$