Question Number 221577 by Nicholas666 last updated on 08/Jun/25
$$ \\ $$$$\:\:\:\:\:\:\mathrm{Prove};\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:\frac{{x}\mathrm{d}{x}}{\left({x}^{\mathrm{2}} \:+\:\mathrm{1}\right)\left({e}^{\mathrm{2}\pi{x}} \:−\:\mathrm{1}\right)}\:=\:\frac{\gamma}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:\: \\ $$$$\:\:\:\:\mathrm{where};\:\gamma\:\:\mathrm{is}\:\mathrm{a}\:\mathrm{Euler}'\mathrm{s}\:\mathrm{Mascheroni}\:\mathrm{constant}\:\:\:\: \\ $$$$ \\ $$
Answered by MrGaster last updated on 08/Jun/25
![=∫_0 ^1 (x/(x^2 +1))((1/(2πx))−(1/2)+(1/π)Σ_(k=1) ^∞ (x/(x^2 +k^2 )))dx =(1/(2π))∫_0 ^1 (1/(x^2 +1))dx−(1/2)∫_(0 ) ^1 (x/(x^2 +1))dx+(1/π)Σ_(k=1) ^∞ ∫_0 ^1 (x^2 /((x^2 +1)(x^2 +k^2 )))dx =(1/(2π))[arctan]_0 ^1 −(1/2)[(1/2)ln(x^2 +1)]_0 ^1 +(1/π)Σ_(k=1) ^∞ ∫_0 ^1 (x^2 /((x^2 +1)(x^2 +k^2 )))dx =(1/(2π))((π/4)−0)−(1/4)(ln 2−ln 1)+(1/π)Σ_(k=1) ^∞ ∫_0 ^1 (x^2 /((x^2 +1)(x^2 +k^2 )))dx =(1/8)−(1/4)ln 2+(1/π)Σ_(k=1) ^∞ ∫_0 ^1 (x^2 /((x^2 +1)(x^2 +k^2 )))dx ∫_0 ^1 (x^2 /((x^2 +1)(x^2 +k^2 )))dx=(1/(k^2 −1))∫_0 ^1 ((1/(x^2 +1))−(k^2 /(x^2 +k^2 )))dx (k≠1) =(1/(k^2 −1))([arctan x]_0 ^1 −k[arctan(x/k)]_0 ^(−1) ] =(1/(k^2 −1))((π/4)−k arctan(1/k)) ∫_0 ^1 (x^2 /((x^2 +1)))dx=∫_0 ^(π/4) tan^2 cos^2 θdθ=∫_0 ^(π/4) sin^2 θdθ=(1/2)∫_(0 ) ^(π/4) (1−cos 2θ)dθ=(1/2)[θ−(1/2)sin 2θ]_0 ^(π/4) =(π/(32)) Σ_(k=2) ^∞ ∫_0 ^1 (x^2 /((x^2 −1)(x^2 −k^2 )))dx=(π/(32))+Σ_(k=2) ^∞ (1/(k^2 −1))((π/4)−arctan(1/k)) Σ_(k=2) ^∞ (1/(k^2 −1))=(1/2)Σ_(k=2) ^∞ ((1/(k−1))−(1/(k+1)))=(1/2)(1+(1/2))=(3/4) arctan(1/k)=(1/2)Σ_(k=2) ^∞ ((1/(k−1))+(1/(k+1)))arctan(1/k) =(1/2)Σ_(k=1) ^∞ ((arctan(1/(k+1)))/k)+(1/2)Σ_(k=3) ^∞ ((arctan(1/(k−1)))/k) =(1/2)Σ_(k=1) ^∞ ((arctan(1/(k+1)))/k)+(1/2)Σ_(k=1) ^∞ ((arctan(1/k))/(k+2)) =(1/2)Σ_(k=1) ^∞ arctan(1/k)((1/k)−(1/(k+2))) =(1/2)Σ_(k=1) ^∞ arctan(1/k)∫_0 ^∞ e^(−t) (e^(−kt) −e^(−(k+2)t) )dt =(1/2)∫_0 ^∞ e^(−t) (1−e^(2t) )Σ_(k=1) ^∞ arctan(1/k)e^(−kt) dt Σ_(k=1) ^∞ arctan(1/k)e^(−kt) =ℑΣ_(k=1) ^∞ Ei(i/k)e^(−kt) =γ(e^(−t) /(1−e^(−t) ))+(π/2) (e^(−t) /(1−e^(−t) ))−ln(1+e^(−2t) ) ∫_0 ^∞ e^(−t) (1−e^(−2t) )(γ(e^(−t) /(1−e^(−t) ))+(π/2) (e^(−t) /(1−e^(−t) ))−(1/2)ln(1+e^(−2t) ) (1/π)Σ_(k=1) ^∞ ∫_(0 ) ^1 (x^2 /((x^2 +1)(x^2 +k^2 )))dx=(1/π)((π/(32))+(π/4)∙(3/4)−(1/2)((γ/2)+(π/8)−(1/4))) =(1/(32))+(3/(16))−(γ/(4π))−(1/(16))+(1/(8π)) =(1/(32))+(6/(32))−(γ/(4π))+(1/(8π))=(7/(32))−(γ/(4π))+(1/(8π)) I=(1/8)−(1/4)ln 2+(7/(32))−(γ/(4π))+(1/(8π)) =(4/(32))+(7/(32))−(1/4)ln 2−(γ/(4π))+(1/(8π))=((11)/(32))−(1/4)ln 2−(γ/(4π))+(1/(8π)) =(γ/2)−(1/4)](https://www.tinkutara.com/question/Q221579.png)
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}\left(\frac{\mathrm{1}}{\mathrm{2}\pi{x}}−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\pi}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{x}}{{x}^{\mathrm{2}} +{k}^{\mathrm{2}} }\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}\:} ^{\mathrm{1}} \frac{{x}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}+\frac{\mathrm{1}}{\pi}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\left[\mathrm{arctan}\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{1}}{\pi}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\pi}\left(\frac{\pi}{\mathrm{4}}−\mathrm{0}\right)−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{ln}\:\mathrm{2}−\mathrm{ln}\:\mathrm{1}\right)+\frac{\mathrm{1}}{\pi}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}+\frac{\mathrm{1}}{\pi}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)}{dx}=\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}−\frac{{k}^{\mathrm{2}} }{{x}^{\mathrm{2}} +{k}^{\mathrm{2}} }\right){dx}\:\left({k}\neq\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{1}}\left(\left[\mathrm{arctan}\:{x}\right]_{\mathrm{0}} ^{\mathrm{1}} −{k}\left[\mathrm{arctan}\frac{{x}}{{k}}\right]_{\mathrm{0}} ^{−\mathrm{1}} \right] \\ $$$$=\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{1}}\left(\frac{\pi}{\mathrm{4}}−{k}\:\mathrm{arctan}\frac{\mathrm{1}}{{k}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)}{dx}=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{tan}^{\mathrm{2}} \mathrm{cos}^{\mathrm{2}} \theta{d}\theta=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \mathrm{sin}^{\mathrm{2}} \theta{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}\:} ^{\pi/\mathrm{4}} \left(\mathrm{1}−\mathrm{cos}\:\mathrm{2}\theta\right){d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\left[\theta−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}\theta\right]_{\mathrm{0}} ^{\pi/\mathrm{4}} =\frac{\pi}{\mathrm{32}} \\ $$$$\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({x}^{\mathrm{2}} −{k}^{\mathrm{2}} \right)}{dx}=\frac{\pi}{\mathrm{32}}+\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{1}}\left(\frac{\pi}{\mathrm{4}}−\mathrm{arctan}\frac{\mathrm{1}}{{k}}\right) \\ $$$$\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} −\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}−\mathrm{1}}−\frac{\mathrm{1}}{{k}+\mathrm{1}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{arctan}\frac{\mathrm{1}}{{k}}=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{2}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}−\mathrm{1}}+\frac{\mathrm{1}}{{k}+\mathrm{1}}\right)\mathrm{arctan}\frac{\mathrm{1}}{{k}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{arctan}\frac{\mathrm{1}}{{k}+\mathrm{1}}}{{k}}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{3}} {\overset{\infty} {\sum}}\frac{\mathrm{arctan}\frac{\mathrm{1}}{{k}−\mathrm{1}}}{{k}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{arctan}\frac{\mathrm{1}}{{k}+\mathrm{1}}}{{k}}+\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{arctan}\frac{\mathrm{1}}{{k}}}{{k}+\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\frac{\mathrm{1}}{{k}}\left(\frac{\mathrm{1}}{{k}}−\frac{\mathrm{1}}{{k}+\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\frac{\mathrm{1}}{{k}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} \left({e}^{−{kt}} −{e}^{−\left({k}+\mathrm{2}\right){t}} \right){dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} \left(\mathrm{1}−{e}^{\mathrm{2}{t}} \right)\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\frac{\mathrm{1}}{{k}}{e}^{−{kt}} {dt} \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{arctan}\frac{\mathrm{1}}{{k}}{e}^{−{kt}} =\Im\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\mathrm{Ei}\left({i}/{k}\right){e}^{−{kt}} \\ $$$$=\gamma\frac{{e}^{−{t}} }{\mathrm{1}−{e}^{−{t}} }+\frac{\pi}{\mathrm{2}}\:\frac{{e}^{−{t}} }{\mathrm{1}−{e}^{−{t}} }−\mathrm{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{t}} \right) \\ $$$$\int_{\mathrm{0}} ^{\infty} {e}^{−{t}} \left(\mathrm{1}−{e}^{−\mathrm{2}{t}} \right)\left(\gamma\frac{{e}^{−{t}} }{\mathrm{1}−{e}^{−{t}} }+\frac{\pi}{\mathrm{2}}\:\frac{{e}^{−{t}} }{\mathrm{1}−{e}^{−{t}} }−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+{e}^{−\mathrm{2}{t}} \right)\right. \\ $$$$\frac{\mathrm{1}}{\pi}\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}\:} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} +{k}^{\mathrm{2}} \right)}{dx}=\frac{\mathrm{1}}{\pi}\left(\frac{\pi}{\mathrm{32}}+\frac{\pi}{\mathrm{4}}\centerdot\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\gamma}{\mathrm{2}}+\frac{\pi}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}\right)\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}+\frac{\mathrm{3}}{\mathrm{16}}−\frac{\gamma}{\mathrm{4}\pi}−\frac{\mathrm{1}}{\mathrm{16}}+\frac{\mathrm{1}}{\mathrm{8}\pi} \\ $$$$=\frac{\mathrm{1}}{\mathrm{32}}+\frac{\mathrm{6}}{\mathrm{32}}−\frac{\gamma}{\mathrm{4}\pi}+\frac{\mathrm{1}}{\mathrm{8}\pi}=\frac{\mathrm{7}}{\mathrm{32}}−\frac{\gamma}{\mathrm{4}\pi}+\frac{\mathrm{1}}{\mathrm{8}\pi} \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}+\frac{\mathrm{7}}{\mathrm{32}}−\frac{\gamma}{\mathrm{4}\pi}+\frac{\mathrm{1}}{\mathrm{8}\pi} \\ $$$$=\frac{\mathrm{4}}{\mathrm{32}}+\frac{\mathrm{7}}{\mathrm{32}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}−\frac{\gamma}{\mathrm{4}\pi}+\frac{\mathrm{1}}{\mathrm{8}\pi}=\frac{\mathrm{11}}{\mathrm{32}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mathrm{2}−\frac{\gamma}{\mathrm{4}\pi}+\frac{\mathrm{1}}{\mathrm{8}\pi} \\ $$$$=\frac{\gamma}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}} \\ $$
Commented by Nicholas666 last updated on 08/Jun/25
$$\mathrm{beautifull}\:\mathrm{solution} \\ $$