According to WolframAlpha, it is equal to approximately $1.35619$ and $\frac{3\pi}{4}-1\approx1.3561944901$.
The integral seems way too difficult for me but I know of the possibly related integral $\int_{-\infty}^{\infty}e^{-x^2}dx=\sqrt{\pi}$ or the other possibly related integral $\int_{0}^{\infty}\ln\left(e^{-x}+1\right)dx=\frac{\pi^{2}}{12}$. I've tried u-subs and integration by parts and have come to dead ends with both. How do I solve this integral?
Best Answer
The actual value is close to but different from $\frac{3\pi}{4}-1$. Integration by parts and $(25.5.3)$ give \begin{align*} \int_{ - \infty }^{ + \infty } {\log (1 + {\rm e}^{ - x^2 } ){\rm d}x} & = 2\int_0^{ + \infty } {\log (1 + {\rm e}^{ - x^2 } ){\rm d}x} \mathop = \limits^{{\rm IBP}} 4\int_0^{ + \infty } {\frac{{x^2 }}{{{\rm e}^{x^2 } + 1}}{\rm d}x} \\ &\!\mathop = \limits^{t = x^2 } 2\int_0^{ + \infty } {\frac{{\sqrt t }}{{{\rm e}^t + 1}}{\rm d}t} = (2 - \sqrt 2 )\Gamma\! \left( {\tfrac{3}{2}} \right)\!\zeta\! \left( {\tfrac{3}{2}} \right) = 1.3561877903 \ldots \end{align*}