级数

\[\sum_{n=0}^\infty \frac{\cos(n\theta)}{x^{n+1}}= \frac{x-\cos \theta}{x^2-2x\cos \theta+1} \tag{1} \]

\[\sum_{n=1}^\infty \frac{\sin(n\theta)}{x^{n+1}}= \frac{\sin \theta}{x^2-2x\cos \theta+1}\tag{2} \]

(1) and (2) are the real and imaginary parts of the geometric series

\[sum_{n=0}^\infty \left( \frac{e^{i\theta}}{x}\right)^n = \frac{x}{x-e^{i\theta}} \]

For suitable (a), we have

\[\int_0^{\infty} \frac{a-\cos(x)}{(a^2-2a\cos x+1)(1+x^2)}dx = \frac{\pi e}{2(ae-1)}\tag{3} \]

\[\int_{0}^{1}\frac{\ln\left ( x+\sqrt{x^{2}+1} \right )}{x}\mathrm{d}x=\int_{0}^{1}\frac{\mathrm{arcsinh} x}{x}\mathrm{d}x \]

\[{\begin{align*} \int_{0}^{1}\frac{\ln\left ( x+\sqrt{x^{2}+1} \right )}{x}\mathrm{d}x&=-\int_{0}^{1}\frac{\ln x}{\sqrt{1+x^{2}}}\mathrm{d}x=-\mathcal{I}'\left ( 0 \right )\\ &=\color{blue}{\mathrm{arcsinh}\left ( 1 \right )-\frac{1}{2}\, _{2}F_{1}^{\left ( 0,1,0,0 \right )}\left ( \frac{1}{2},\frac{1}{2};\frac{3}{2};-1 \right )}\\[8pt] &~~~\color{blue}{-\frac{1}{2}\, _{2}F_{1}^{\left ( 0,0,1,0 \right )}\left ( \frac{1}{2},\frac{1}{2};\frac{3}{2};-1 \right )}\\[8pt] &\approx 0.95520180648118 \end{align*}}\]

\[\[\int_0^\infty \frac{a-\cos (x)}{a^2-2a \cos x +1}\cdot \frac{1}{1+x^4}dx=\frac{\pi}{2\sqrt{2}}\exp \left( \frac{1}{\sqrt{2}}\right)\frac{a \exp \left(\frac{1}{\sqrt{2}} \right)+\sin \left( \frac{1}{\sqrt{2}}\right)-\cos \left( \frac{1}{\sqrt{2}}\right)}{1-2a \exp \left(\frac{1}{\sqrt{2}}\right)\cos \left( \frac{1}{\sqrt{2}}\right) +a^2 \exp \left(\sqrt{2}\right)} \tag{4} \]

\[\Large\displaystyle \int^{\infty}_{0}\frac{\tanh\left(\, x\,\right)} {x\left[\, 1 - 2\cosh\left(\, 2x\,\right)\,\right]^{2}}\,{\rm d}x\]

Solution:Solution:
A possible way I see of doing this is to apply the substitution x↦−lnxx↦−ln⁡x, which yields

\[\begin{align*} \mathcal{I} &=-\int_{0}^{1}\frac{x^3(1-x^2)}{(1+x^2)(1-x^2+x^4)^2}\frac{\mathrm{d}x}{\ln{x}}\\&=-\int_{0}^{1}\frac{x^3(1-x^2)}{(1+x^6)(1-x^2+x^4)}\frac{\mathrm{d}x}{\ln{x}}\\ &=-\int_{0}^{1}\frac{x^3(1-x^4)}{(1+x^6)^2}\frac{\mathrm{d}x}{\ln{x}}\\&=-\int_{0}^{1}\frac{z(1-z^{4/3})}{(1+z^2)^2}\frac{\mathrm{d}z}{3z^{2/3}\ln{\left(z^{1/3}\right)}}\\&=\int_{0}^{1}\,\frac{1}{(1+z^2)^2}\frac{z^{5/3}-z^{1/3}}{\ln{z}}\mathrm{d}z\\ &=\int_{0}^{1}\,\frac{\mathrm{d}z}{(1+z^2)^2}\int_{1/3}^{5/3}\,z^{\mu}\mathrm{d}\mu\\&=\int_{1/3}^{5/3}\mathrm{d}\mu\int_{0}^{1}\,\frac{z^{\mu}}{(1+z^2)^2}\mathrm{d}z\\&=\int_{1/3}^{5/3}\left[-\frac14+\frac{\mu-1}{4}\beta{\left(\frac{\mu-1}{2}\right)}\right]\mathrm{d}\mu\\ &=-\frac13+\int_{1/3}^{5/3}\left[\frac{\mu-1}{4}\beta{\left(\frac{\mu-1}{2}\right)}\right]\mathrm{d}\mu\\&=-\frac13+\int_{-1/3}^{1/3}\,t\beta{\left(t\right)}\mathrm{d}t\\ &=-\frac13+\int_{-1/3}^{1/3}\,\frac{t}{2}\left[\psi{\left(\frac{t+1}{2}\right)}-\psi{\left(\frac{t}{2}\right)}\right]\mathrm{d}t\\&=-\frac13+\int_{-1/3}^{1/3}\,\frac{t}{2}\psi{\left(\frac{t+1}{2}\right)}\mathrm{d}t-\int_{-1/3}^{1/3}\,\frac{t}{2}\psi{\left(\frac{t}{2}\right)}\mathrm{d}t\\ &=-\frac13+\int_{1/3}^{2/3}\,(2u-1)\psi{\left(u\right)}\mathrm{d}u-2\int_{-1/6}^{1/6}\,u\psi{\left(u\right)}\mathrm{d}u\\ &=-\frac13-\int_{1/3}^{2/3}\,\psi{\left(u\right)}\mathrm{d}u+2\int_{1/3}^{2/3}\,u\psi{\left(u\right)}\mathrm{d}u-2\int_{-1/6}^{1/6}\,u\psi{\left(u\right)}\mathrm{d}u\\ &=-\frac13+\ln{\left(\frac{\Gamma{\left(\dfrac13\right)}}{\Gamma{\left(\dfrac23\right)}}\right)}+2\int_{1/3}^{2/3}\,u\psi{\left(u\right)}\mathrm{d}u-2\int_{-1/6}^{1/6}\,u\psi{\left(u\right)}\mathrm{d}u\\ &=-\frac13+\ln{\left(\frac{\Gamma{\left(\dfrac13\right)}}{\Gamma{\left(\dfrac23\right)}}\right)}+2\int_{1/3}^{2/3}\,u\psi{\left(u\right)}\mathrm{d}u-2\int_{5/6}^{7/6}\,(1-v)\psi{\left(1-v\right)}\mathrm{d}v\\ &=-\frac13+\ln{\left(\frac{\Gamma{\left(\dfrac13\right)}}{\Gamma{\left(\dfrac23\right)}}\right)}+2\left[u\ln{\Gamma\left(u\right)}-\psi^{(-2)}{\left(u\right)}\right]_{1/3}^{2/3}\\ &~~~~~ +2\left[(1-v)\ln{\Gamma\left(1-v\right)}-\psi^{(-2)}{\left(1-v\right)}\right]_{5/6}^{7/6}\\ &=\ln{\left(\frac{\Gamma{\left(\dfrac13\right)}}{\Gamma{\left(\dfrac23\right)}}\right)}-\frac{5\pi}{9\sqrt{3}}-\(\frac{\Gamma{\left(\dfrac23\right)}^2}{\Gamma{\left(\dfrac13\right)}}\right)}+\frac{5\psi^{(1)}{\left(\dfrac13\right)}}{6\sqrt{3}\,\pi}\\ &=\Large\boxed{\displaystyle\color{blue}{-\frac{5\pi}{9\sqrt{3}}-\frac{\ln{3}}{6}+\frac{5\psi^{(1)}{\left(\dfrac13\right)}}{6\sqrt{3}\,\pi}}} \end{align*}\]