Find $\lambda$ if $\int^{\infty}_0 \frac{\log(1+x^2)}{(1+x^2)}dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}dx$

Question

Problem : If $\displaystyle\int^\infty_0 \frac{\log(1+x^2)}{(1+x^2)}\,dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}\,dx$ then find the value of $\lambda$.

I am not getting any clue how to proceed as if I put $(1+x^2)\,dx =t $ then its derivative is not available. Please suggest how to proceed in this. Thanks.

Answer 1

Setting $x=\tan y,$

$$I=\int_0^\infty\frac{\ln(1+x^2)}{1+x^2}\ dx=\int_0^{\dfrac\pi2}\ln(\sec^2y)\ dy=-2\int_0^{\dfrac\pi2}\ln(\cos y)\ dy (\text{ as } \cos y\ge0 \text{ here})$$

which is available here : Evaluate $\int_0^{\pi/2}\log\cos(x)\,\mathrm{d}x$

The Right Hand Side can be found here : Evaluate the integral: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$

Answer 2

From Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$, you can obtain $$\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx=\pi\ln 2. $$ Define $$I(\alpha)=\int_0^{1}\frac{\ln(\alpha x+1)}{x^2+1}dx. $$ Then \begin{eqnarray*} I'(\alpha)&=&\int_0^{1}\frac{x}{(\alpha x+1)(x^2+1)}dx=\int_0^1\left(\frac{x+\alpha}{(\alpha^2+1)(x^2+1)}-\frac{\alpha}{(\alpha^2+1)(x^2+1)}\right)dx.\\ &=&\frac{\pi\alpha+2\ln 2-4\ln(\alpha+1)}{4(\alpha^2+1)}. \end{eqnarray*} Hence \begin{eqnarray*} I(1)&=&\int_0^1\frac{\pi\alpha+2\ln 2-4\ln(\alpha+1)}{4(\alpha^2+1)}d\alpha\\ &=&\int_0^1\frac{\pi\alpha+2\ln 2}{4(\alpha^2+1)}d\alpha-I(1) \end{eqnarray*} and so $$ I(1)=\frac{1}{2} \int_0^1\frac{\pi\alpha+2\ln 2}{4(\alpha^2+1)}d\alpha=\frac{1}{8}\pi\ln 2. $$ Thus $\lambda=8$.