Cómo calcular $\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$

Question

Primero he escrito $x^4-2x^2+2=(x^2-1)^2+1$ entonces elijo $t=x^2-1$ . De esto obtuve $$\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt$$

¿A partir de aquí cómo proceder?

Answer 1

CONSEJO :

$$\frac{\text{d}}{\text{dn}}\left(\int_0^\infty\frac{\ln\left(1+\text{n}x^2\right)}{\sqrt{x}\left(1+x\right)}\space\text{d}x\right)=\int_0^\infty\frac{x^\frac{3}{2}}{\left(1+\text{n}x^2\right)\left(1+x\right)}\space\text{d}x=$$ $$\frac{\pi}{2\left(1+\text{n}\right)}\left(2-\frac{\sqrt{2}\left(\sqrt{\text{n}}-1\right)}{\text{n}^\frac{3}{4}}\right)\tag1$$

Answer 2

Sea $$\begin{eqnarray} I(a)=\int_0^\infty\frac{\log(1+ax^4)}{1+x^2}dx \end{eqnarray}$$ Entonces, $$\begin{eqnarray} I'(a)&=&\int_0^\infty \frac{x^4}{(1+ax^4)(1+x^2)}dx\\ &=&\frac{1}{1+a}\int_0^\infty\left(\frac{1}{1+x^2}+\frac{x^2-1}{1+a x^4}\right)dx =\frac{\pi}{1+a}\left(\frac12+\frac{1-a^{1/2}}{2\sqrt2a^{3/4}}\right)\\ \end{eqnarray}$$ Así, $$\begin{align} &\frac{1}{2}\int\limits_{0}^{\infty}\frac{\ln{(1+t^2)}}{\sqrt{t}(1+t)}dt \overset{x=\sqrt t}=\int_0^\infty\frac{\log(1+ax^4)}{1+x^2}dx\\ = & I(1)=\int_0^1 I’(\alpha)d\alpha = \int_0^1 \frac{\pi}{1+a}\left(\frac12+\frac{1-a^{1/2}}{2\sqrt2a^{3/4}}\right)d\alpha \\ = & \frac\pi2(\ln2 + 2\coth^{-1} \sqrt2 ) \end{align}$$