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There is an interesting trick that can be used to evaluate integrals in the form $$I=\int_{-a}^a \frac{E(x)}{b^x+1}dx$$ where $E$ is an even function. Notice that, by substituting $x\to -x$, $$I=\int_{-a}^a \frac{E(-x)}{b^{-x}+1}dx=\int_{-a}^a \frac{b^xE(x)}{b^{x}+1}dx$$ and so $$I+I=\int_{-a}^a \frac{E(x)+b^xE(x)}{b^x+1}dx=\int_{-a}^a E(x)dx$$ and so $$I=\frac{1}{2}\int_{-a}^a E(x)dx=\int_{0}^a E(x)dx$$ For example, this trick can be used to evaluate the intimidating integral $$\int_{-1}^1 \frac{x^{100}}{e^x+1}dx=\frac{1}{101}$$

QUESTION: Is there some way to generalize this trick to integrals of the form $$I=\int_{-a}^a \frac{E(x)}{(b^x+1)^2}dx$$ or will this type of integral just have to be done the hard way?

Franklin Pezzuti Dyer
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  • I have my doubts about your method You are claiming that $\int_{-a}^{a} \frac{E(x)}{b^x+1}dx = \int_{-a}^{a} \frac{E(y)}{b^y+1}dy$. Which is all well and good, but if you next say that $y=-x$ then $dy=-dx$ thus an extra minus sign appears and the nice canceling does not occur. – Daniel Gendin Mar 02 '18 at 00:30
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    @DanielGendin When $x\to -x$, the bounds change from $-a$ and $a$ to $a$ and $-a$. This "extra minus sign" changes the order of the bounds back to $-a$ and $a$ for me, so no harm is done. – Franklin Pezzuti Dyer Mar 02 '18 at 00:31
  • you are right, thanks for clarifying – Daniel Gendin Mar 02 '18 at 17:35

1 Answers1

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Lets $\ds{{1 \over b^{x} + 1} = {1 \over \expo{\beta x} + 1} \equiv \mathrm{f}\pars{x}:\ Fermi\!-\!Dirac\ Distribution\ Function.\quad\beta \equiv \ln\pars{b} > 0}$

\begin{align} &\mbox{It satisfies the identity}\quad \mrm{f}\pars{x} = \left\{\begin{array}{lcrcl} \ds{\Theta\pars{-x} + \,\mrm{sgn}\pars{x}\mrm{f}\pars{\verts{x}}} & \mbox{if} & \ds{x} & \ds{\not=} & \ds{0} \\[2mm] \ds{1 \over 2} & \mbox{if} & \ds{x} & \ds{=} & \ds{0} \end{array}\right. \\[5mm] &\mbox{where}\ \Theta\ \mbox{is the}\ Heaviside\ Step\ Function. \end{align}

Then, with $\ds{a > 0}$: $$ \bbx{\int_{-a}^{a}\mrm{E}\pars{x}\mrm{f}\pars{x}\dd x = \int_{-a}^{0}\mrm{E}\pars{x}\dd x + \int_{0}^{a}\bracks{\mrm{E}\pars{x} - \mrm{E}\pars{-x}}\mrm{f}\pars{x}\dd x} $$

Moreover

\begin{align} \int_{-a}^{a}{\mrm{E}\pars{x} \over \pars{b^{x} + 1}^{2}}\,\dd x & = \int_{-a}^{a}\mrm{E}\pars{x} \bracks{-\,{1 \over \beta}\expo{-\beta x}\mrm{f}'\pars{x}}\,\dd x \\[1cm] & \stackrel{\mrm{IBP}}{=}\,\,\, -\,{\mrm{E}\pars{a}\expo{-\beta a}\mrm{f}\pars{a} - \mrm{E}\pars{-a}\expo{\beta a}\mrm{f}\pars{-a} \over \beta} \\[2mm] & + \int_{-a}^{a}\bracks{% {1 \over \beta}\,\mrm{E}'\pars{x}\expo{-\beta x} - \mrm{E}\pars{x}\expo{-\beta x}}\mrm{f}\pars{x}\,\dd x \end{align}

which is reduced to the previous case !!!.

Felix Marin
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