Improper integral involving $e^x$

Question

How do you show that $$\int_{-\infty }^{+\infty }{\frac{{{x}^{2}}\, \text{d}x}{\left( \beta +{{\text{e}}^{x}} \right)\left( 1-{{\text{e}}^{-x}} \right)}}=\frac{\left( {{\pi }^{2}}+{{\ln }^{2}}\beta \right)\ln \beta }{3\left( \beta +1 \right)},\ \ \left|\text{arg} \ \beta\right|<\pi ?$$

Answer 1

Let $\displaystyle u = e^{x}$.

Then $$\int_{-\infty}^{\infty} \frac{x^{2}}{(\beta+e^x)(1-e^{-x})} \ dx = \int_{0}^{\infty} \frac{\ln^{2} u}{(\beta + u)(u-1)} \ du . $$

Now let $\displaystyle f(z) = \frac{\ln^{3} z}{(\beta+z)(z-1)}$ and integrate around a keyhole contour where the branch cut for $\log z$ is placed along the positive real axis.

Notice that there is a simple pole at $z=1$ below the branch cut.

So the contour needs to be indented around $z=1$ below the branch cut.

Then

$$\int_{0}^{\infty} \frac{\ln^3 u}{(\beta+u)(u-1)} \ du -\pi i \ \text{Res}[f(z), e^{2 \pi i}] - \text{PV} \int_{0}^{\infty} \frac{(\ln u +2 \pi i)^{3}}{(\beta+u)(u-1)} \ du = 2 \pi i \ \text{Res} [f,-\beta]$$

where $$ \text{Res}[f(z), - \beta] = \ \frac{\ln^{3}(-\beta)}{-\beta - 1} = - \frac{(\ln \beta + i \pi)^{3}}{\beta + 1}$$

and

$$ \text{Res}[f(z), e^{2 \pi i} ] = -\frac{8 i \pi^{3}}{\beta + 1} .$$

Equating imaginary parts on both sides of the equation,

$$ \text{PV} \displaystyle \int_{0}^{\infty} \frac{-6 \pi \ln^{2} u + 8 \pi^{3}}{(\beta+u)(u-1)} \ du = \frac{- 2 \pi \ln^{3} \beta + 6 \pi^{3} \ln \beta}{\beta+1} .$$

Then rearranging,

$$ \begin{align} \int_{0}^{\infty} \frac{\ln^{2}u}{(\beta + u)(u-1)} \ du &= \frac{ \ln^{3} \beta - 3 \pi^{2} \ln \beta}{3(\beta+1)} + \frac{4}{3} \pi^{2} \ \text{PV} \int_{0}^{\infty} \frac{1}{(\beta +u)(u-1)} \ du \\ &= \frac{\ln^{3} \beta - 3 \pi^{2} \ln \beta}{3(\beta+1)} + \frac{4 \pi^{2}}{3} \frac{\ln \beta}{\beta +1} \\ &= \frac{(\pi^2 + \ln^{2} \beta) \ln \beta}{3(\beta+1)} . \end{align}$$

EDIT:

$$ \begin{align} \text{PV} \int_{0}^{\infty} \frac{1}{(\beta +u)(u-1)} \ du &= \lim_{\epsilon \to 0^{+}} \Bigg( \int_{0}^{1-\epsilon} \frac{du}{(\beta +u)(u-1)} + \int_{1+\epsilon}^{\infty} \frac{du}{(\beta +u)(u-1)} \Bigg) \\ &= \frac{1}{\beta +1} \lim_{\epsilon \to 0^{+}} \Bigg( \ln \Big(\frac{u-1}{\beta + u}\Big) \Big|^{1- \epsilon}_{0} + \ln \Big(\frac{u-1}{\beta + u}\Big) \Big|^{\infty}_{1+\epsilon}\Bigg) \\ &= \frac{1}{\beta+1} \lim_{\epsilon \to 0^{+}} \Bigg( \ln \Big(\frac{-\beta-1-\epsilon}{\beta+1-\epsilon} \Big) - \ln \big(\frac{-1}{\beta} \big) \Bigg) \\ &= \frac{\ln (-1) - \ln \big( \frac{-1}{\beta}\big) }{\beta+1} \\ &= \frac{i\pi + \ln \beta - i \pi}{1+ \beta} = \frac{\ln \beta}{\beta+1}. \end{align}$$

SECOND EDIT:

And equating real parts on both sides of the equation, $$ \int_{0}^{\infty} \frac{\log u}{(\beta +u)(u-1)} \ du = \frac{1}{2(\beta +1)} \left(\pi^{2}+\log^{2}(\beta) \right) .$$

Initially I forgot to include the contribution from the indentation around $z=1$ below the branch cut. It didn't matter when equating imaginary parts on both sides of the equation.

Answer 2

Let $u=e^{x}$, $du=u dx$ and the integral becomes

$$I = \int_0^{\infty} du \: \frac{\log^2{u}}{(\beta+u)(u-1)} $$

To evaluate this via the Residue Theorem, consider the integral

$$\oint_C dz \frac{\log^3{z}}{(\beta+z)(z-1)} $$

where $C$ is the standard "keyhole" contour that encircles the origin outside of the positive real axis, and traverses a straight line above and below the real axis. I will skip the full analysis of this contour for now, and just state that

$$\int_0^{\infty} du \: \frac{\log^3{u}}{(\beta+u)(u-1)} - \int_0^{\infty} du \: \frac{\log^3{( u e^{i 2 \pi)}}}{(\beta+u)(u-1)} = i 2 \pi \mathrm{Res}_{z=-\beta} \frac{\log^3{z}}{(\beta+z)(z-1)}$$

or

$$\int_0^{\infty} du \: \frac{\log^3{u} - (\log^3{u} + i 2 \pi)^3}{(\beta+u)(u-1)} = -i 2 \pi \frac{(\log{\beta}+i \pi)^3}{1+\beta} $$

Consider the LHS, which simplifies to

$$\int_0^{\infty} du \: \frac{-i 6 \pi \log^2{u} + 12 \pi^2 \log{u} + i 8 \pi^3 }{(\beta+u)(u-1)} = -i 2 \pi \left (I - 4 \pi^2 \int_0^{\infty} du \: \frac{1}{(\beta+u)(u-1)}\right ) $$

The integral in the parentheses is the Cauchy Principal Value:

$$PV\int_0^{\infty} du \: \frac{1}{(\beta+u)(u-1)} = \frac{\log{\beta}}{1+\beta}$$

Dividing by $-i 2 \pi$, equating real and imaginary parts above, and taking the real part, we get

$$3 I-4 \pi^2 \frac{\log{\beta}}{1+\beta} = \frac{\log^3{\beta} - 3 \pi^2\log{\beta}}{1+\beta}$$

and the stated result follows. PV computation verified by @RandomVariable above.