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Display information for equation id:math.2555.51 on revision:2555

* Page found: Das ideale Fermigas (eq math.2555.51)

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TeX (original user input): 

\begin{align}

& I=\int_{-\infty }^{\infty }{{}}dx{{x}^{2}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}=2\int_{0}^{\infty }{{}}dx{{x}^{2}}\frac{{{e}^{x}}}{{{\left( {{e}^{x}}+1 \right)}^{2}}}=-2\left[ {{x}^{2}}\frac{1}{\left( {{e}^{x}}+1 \right)} \right]_{0}^{\infty }+4\int_{0}^{\infty }{{}}dx\frac{x}{\left( {{e}^{x}}+1 \right)} \\

& \left[ {{x}^{2}}\frac{1}{\left( {{e}^{x}}+1 \right)} \right]_{0}^{\infty }=0 \\

& \int_{0}^{\infty }{{}}dx\frac{x}{\left( {{e}^{x}}+1 \right)}=\frac{{{\pi }^{2}}}{12} \\

& \Rightarrow I=\frac{{{\pi }^{2}}}{3} \\

\end{align}

TeX (checked): 

{\begin{aligned}&I=\int _{-\infty }^{\infty }{}dx{{x}^{2}}{\frac {{e}^{x}}{{\left({{e}^{x}}+1\right)}^{2}}}=2\int _{0}^{\infty }{}dx{{x}^{2}}{\frac {{e}^{x}}{{\left({{e}^{x}}+1\right)}^{2}}}=-2\left[{{x}^{2}}{\frac {1}{\left({{e}^{x}}+1\right)}}\right]_{0}^{\infty }+4\int _{0}^{\infty }{}dx{\frac {x}{\left({{e}^{x}}+1\right)}}\\&\left[{{x}^{2}}{\frac {1}{\left({{e}^{x}}+1\right)}}\right]_{0}^{\infty }=0\\&\int _{0}^{\infty }{}dx{\frac {x}{\left({{e}^{x}}+1\right)}}={\frac {{\pi }^{2}}{12}}\\&\Rightarrow I={\frac {{\pi }^{2}}{3}}\\\end{aligned}}

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I = ∫ - ∞ ∞ 𝑑 x ⁢ x 2 ⁢ e x ( e x + 1 ) 2 = 2 ⁢ ∫ 0 ∞ 𝑑 x ⁢ x 2 ⁢ e x ( e x + 1 ) 2 = - 2 ⁢ [ x 2 ⁢ 1 ( e x + 1 ) ] 0 ∞ + 4 ⁢ ∫ 0 ∞ 𝑑 x ⁢ x ( e x + 1 ) [ x 2 ⁢ 1 ( e x + 1 ) ] 0 ∞ = 0 ∫ 0 ∞ 𝑑 x ⁢ x ( e x + 1 ) = π 2 12 ⇒ I = π 2 3 absent 𝐼 superscript subscript differential-d 𝑥 superscript 𝑥 2 superscript 𝑒 𝑥 superscript superscript 𝑒 𝑥 1 2 2 superscript subscript 0 differential-d 𝑥 superscript 𝑥 2 superscript 𝑒 𝑥 superscript superscript 𝑒 𝑥 1 2 2 superscript subscript delimited-[] superscript 𝑥 2 1 superscript 𝑒 𝑥 1 0 4 superscript subscript 0 differential-d 𝑥 𝑥 superscript 𝑒 𝑥 1 absent superscript subscript delimited-[] superscript 𝑥 2 1 superscript 𝑒 𝑥 1 0 0 absent superscript subscript 0 differential-d 𝑥 𝑥 superscript 𝑒 𝑥 1 superscript 𝜋 2 12 absent ⇒ absent 𝐼 superscript 𝜋 2 3 {\displaystyle{\displaystyle\begin{aligned} \par&\displaystyle I=\int_{-\infty% }^{\infty}{{}}dx{{x}^{2}}\frac{{{e}^{x}}}{{{\left({{e}^{x}}+1\right)}^{2}}}=2% \int_{0}^{\infty}{{}}dx{{x}^{2}}\frac{{{e}^{x}}}{{{\left({{e}^{x}}+1\right)}^{% 2}}}=-2\left[{{x}^{2}}\frac{1}{\left({{e}^{x}}+1\right)}\right]_{0}^{\infty}+4% \int_{0}^{\infty}{{}}dx\frac{x}{\left({{e}^{x}}+1\right)}\\ \par&\displaystyle\left[{{x}^{2}}\frac{1}{\left({{e}^{x}}+1\right)}\right]_{0}% ^{\infty}=0\\ \par&\displaystyle\int_{0}^{\infty}{{}}dx\frac{x}{\left({{e}^{x}}+1\right)}=% \frac{{{\pi}^{2}}}{12}\\ \par&\displaystyle\Rightarrow I=\frac{{{\pi}^{2}}}{3}\\ \par\end{aligned}}}

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I=∫−∞∞dxx2ex(ex+1)2=2∫0∞dxx2ex(ex+1)2=−2[x21(ex+1)]0∞+4∫0∞dxx(ex+1)[x21(ex+1)]0∞=0∫0∞dxx(ex+1)=π212⇒I=π23

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