Does $\int_{-\infty}^{\infty}\ln\left(e^{-x^{2}}+1\right)dx=\frac{3\pi}{4}-1$?

Question

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?

Answer 1

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*}

Answer 2

\begin{align} \int_{-\infty}^{\infty} \ln \left(1 + e^{-x^2}\right)\mathrm d x &= \sum_{n = 1}^{\infty} \frac{(-1)^{n+1}}{n} \int_{-\infty}^{\infty} e^{-nx^2}\mathrm dx\\ &= \sum_{n=1}^{\infty} \sqrt{\pi} \frac{(-1)^{n+1}}{n\sqrt n}\\ &= \sqrt{\pi} \eta\left(\frac32\right) \end{align}

Where $\eta$ is the Dirichlet eta function.

Answer 3

By integration by parts and infinite series, we have $$ \begin{aligned} I&= 2\left[x \ln \left(e^{-x^2}+1\right)\right]_0^{\infty}-2 \int_{-\infty}^{\infty} \frac{x \cdot(-2 x) e^{-x^2}}{e^{-x^2}+1} d x\\&= 4 \int_0^{\infty} \frac{x^2 e^{-x^2}}{e^{-x^2}+1} d x \\ & =4 \sum_{n=1}^{\infty}(-1)^{n-1} \int_0^{\infty} x^2 e^{-n x^2} d x \end{aligned} $$ Via integration by parts again on the last integral, we get $$ \begin{aligned} \int_0^{\infty} x^2 e^{-n x^2} d x = & -\frac{1}{2 n} \int_0^{\infty} x d\left(e^{-n x^2}\right) \\ = & \frac{1}{2 n} \int_0^{\infty} e^{-(\sqrt{n} x)^2} d x \\= & \frac{\sqrt{\pi}}{4 n^{\frac{3}{2}}} \end{aligned} $$ We can now conclude that $$ \begin{aligned} I & =\sqrt{\pi} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{\frac{3}{2}}} \\ & =\sqrt{\pi} \eta\left(\frac{3}{2}\right) \\ OR & =\frac{\sqrt{\pi}}{2}(\sqrt{2}-2) \zeta\left(\frac{3}{2}\right) \end{aligned} $$ Where $\eta$ is the Dirichlet eta function and $\zeta$ is the Zeta function.