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

Question

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

I am not getting any clue. Please suggest how to proceed. Thanks.

Answer 1

$\lambda=8$, as seen below \begin{align} I=&\int^\infty_0\frac{\ln(1+x^2)}{1+x^2}\,dx \overset{x\to\frac{1-x^2}{2x}}= 4\int_0^1 \frac{\ln\frac{1+x^2}{2x}}{1+x^2}dx\\ =& \ 2\int_0^1 \bigg[{\ln\overset{x\to \frac1x}{\frac{1+x^2}{x^2}}+2 \ln \overset{x\to \frac{1-x}{1+x}}{ \frac{1+x^2}{2}}-\ln(1+x^2)}\bigg]\frac{dx}{1+x^2}\\ =& \ 2 I - 8\int^1_0\frac{\ln(1+x)}{1+x^2}\,dx= 8\int^1_0\frac{\ln(1+x)}{1+x^2}\,dx \end{align}

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 $\lambda=8$.

Answer 3

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: $\int_{0}^{1} \frac{\ln(x+1)}{x^2+1} \mathrm dx$