One of the problem solution is as below:
\begin{align}H(x)&=-\int^{\infty}_{-\infty} \frac{1}{2\lambda}e^{\tfrac{{-|x|}}{\lambda}} \ln\biggl(\frac{1}{2\lambda}e^{-\tfrac{|x|}{\lambda}}\biggr)\,dx \tag{1} \\ &= -\ln\Bigl(\frac{1}{2\lambda}\Bigr) \int^{\infty}_{-\infty}\frac{1}{2\lambda}e^{\tfrac{{-|x|}}{\lambda}}\,dx+\frac{1}{\lambda}\int^{\infty}_{-\infty} |x|\frac{1}{2\lambda}e^{\tfrac{{-|x|}}{\lambda}}\,dx$ \tag{2} \\ &= \ln(2\lambda)+\frac{1}{\lambda}\biggl[\int^{0}_{-\infty}-x\frac{1}{2\lambda}e^{\tfrac{x}{\lambda}}\,dx+\int^{\infty}_{0}x\frac{1}{2\lambda}e^{\tfrac{-x}{\lambda}}\,dx\biggr] \tag{3} \\ &= \ln(2\lambda)+\frac{1}{2\lambda}\lambda+\frac{1}{2\lambda}\lambda \tag{4}\\ &=1+\ln(2\lambda) \end{align} And I am confused that how (1) becomes (2),can anyone tell me more specifics about it?
Basically it's just logarithm laws: $$\ln\Bigl(\frac{1}{2\lambda}e^{-|x|/\lambda}\Bigr) =\ln\Bigl(\frac{1}{2\lambda}\Bigr)+\ln\Bigl(e^{-|x|/\lambda}\Bigr) =\ln\Bigl(\frac{1}{2\lambda}\Bigr)-\frac{|x|}{\lambda}$$ and then split the integral. However there is a sign error.
Just guessing, but maybe line (1) should have a negative sign in it.