calculate-0-x-e-x-2-log-1-e-x-dx-

Question Number 143860 by mathmax by abdo last updated on 19/Jun/21

calculate \int_{0}^{\infty} x e^{- x^{2}} \log (1 + e^{x}) dx

Answered by mathmax by abdo last updated on 20/Jun/21

Φ = \int_{0}^{\infty} {xe}^{- x^{2}} \log (1 + e^{x}) dx \Rightarrow Φ = {[- \frac{1}{2} e^{- x^{2}} \log (1 + e^{x})]}_{0}^{\infty}

- \int_{0}^{\infty} - \frac{1}{2} e^{- x^{2}} \times \frac{e^{x}}{1 + e^{x}} dx = \frac{\log 2}{2} + \frac{1}{2} \int_{0}^{\infty} \frac{e^{- x^{2}}}{1 + e^{- x}} dx but

\int_{0}^{\infty} \frac{e^{- x^{2}}}{1 + e^{- x}} dx = \int_{0}^{\infty} e^{- x^{2}} \sum_{n = 0}^{\infty} e^{- nx} dx

= \sum_{n = 0}^{\infty} \int_{0}^{\infty} e^{- x^{2} - nx} dx = \sum_{n = 0}^{\infty} \int_{0}^{\infty} e^{- (x^{2} + 2 \frac{n}{2} x + \frac{n^{2}}{4} - \frac{n^{2}}{4})} dx

= \sum_{n = 0}^{\infty} e^{\frac{n^{2}}{4}} \int_{0}^{\infty} e^{- {(x + \frac{n}{2})}^{2}} dx = \sum_{n = 0}^{\infty} e^{\frac{n^{2}}{4}} \int_{\frac{n}{2}}^{\infty} e^{- z^{2}} dz

if we put \erf (λ) = \int_{λ}^{\infty} e^{- z^{2}} dz we get

\int_{0}^{\infty} \frac{e^{- x^{2}}}{1 + e^{- x}} dx = \sum_{n = 0}^{\infty} e^{\frac{n^{2}}{4}} \erf (\frac{n}{2}) \Rightarrow

Φ = \frac{\log 2}{2} + \frac{1}{2} \sum_{n = 0}^{\infty} \erf (\frac{n}{2}) e^{\frac{n^{2}}{4}}