Evaluate $\int_{0}^{\infty}\dfrac{\mathrm dx}{(e^{\pi x}+e^{-\pi x})(16+x^2)}$

Find the integral

$$I=\int_{0}^{\infty}\dfrac{1}{(e^{\pi x}+e^{-\pi x})(16+x^2)}dx$$

My try:let $x=-t$

$$I=\int_{-\infty}^{0}\dfrac{1}{(e^{\pi x}+e^{-\pi x})(16+x^2)}dx$$ so $$2I=\int_{-\infty}^{\infty}\dfrac{1}{(e^{\pi x}+e^{-\pi x})(16+x^2)}dx$$

Then I can't,Thank you

Solution 1:

You can use Parseval's theorem, which states that, given two functions $f(x)$ and $g(x)$, each having Fourier transforms $\hat{f}(k)$ and $\hat{g}(k)$, respectively (and satisfying certain integrability conditions that are satisfied here), the following relation holds:

$$\int_{-\infty}^{\infty} dx \, f(x) g^*(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, \hat{f}(k) \, \hat{g}^*(k)$$

where * is complex conjugation. Here, $f(x) = \text{sech}{(\pi x)}$ and $g(x) = (16+x^2)^{-1}$. The FT $\hat{g}(k)$ is very well known and I will not go into detail here about it:

$$\hat{g}(k) = \frac{\pi}{4} e^{-4 |k|}$$

The FT $\hat{f}(k)$, on the other hand, is not as well known, but should be, as it is form-preserving. To maintain the flow of this derivation, I will hold off on the derivation of the following until later:

$$\hat{f}(k) = \text{sech}{(k/2)}$$

Let $I$ be the original integral. Obviously, the integrand of $I$ is even and we may therefore extend the integration interval to the entire real line. Thus

$$\begin{align}I &= \frac14 \int_{-\infty}^{\infty} dx \,\frac{\text{sech}{(\pi x)}}{16+x^2}\\ &= \frac{1}{8 \pi} \int_{-\infty}^{\infty} dk \, \text{sech}{\frac{k}{2}} \frac{\pi}{4} e^{-4 |k|}\\ &=\frac{1}{16} \int_{-\infty}^{\infty} dk \frac{e^{-4 |k|}}{e^{k/2}+e^{-k/2}} \\ &= \frac18 \int_0^{\infty} dk \frac{e^{-9 k/2}}{1+e^{-k}}\\&=\frac18 \sum_{m=0}^{\infty} \frac{(-1)^m}{m+\frac{9}{2}}\end{align}$$

Let

$$S = \sum_{m=0}^{\infty} \frac{(-1)^m}{m+\frac{9}{2}}$$

We may evaluate this sum using the residue theorem. Note that

$$\sum_{m=-\infty}^{\infty} \frac{(-1)^m}{m+\frac{9}{2}} = 2 S + 4 \left (1-\frac13+\frac15-\frac17\right)$$

By the residue theorem, the sum on the left is simply

$$-\operatorname*{Res}_{z=-9/2}[ \pi \, \csc{\pi z}] = \pi$$

Therefore the integral is

$$\int_0^{\infty} \frac{dx}{\left (e^{\pi x}+e^{-\pi x}\right) (16+x^2)} = \frac{S}{8} = \frac{\pi}{16}-\frac{19}{105} \approx 0.015397$$

ADDENDUM

The FT of $\text{sech}{(\pi x)}$ may also be derived using the residue theorem. We simply set up the integral as usual and comvert it into a sum as follows:

$$\begin{align}\int_{-\infty}^{\infty} dx \, \text{sech}{(\pi x)} \, e^{i k x} &= 2 \int_{-\infty}^{\infty} dx \frac{e^{i k x}}{e^{\pi x}+e^{-\pi x}}\\ &= 2 \int_{-\infty}^0 dx \frac{e^{i k x}}{e^{\pi x}+e^{-\pi x}} + 2 \int_0^{\infty}dx \frac{e^{i k x}}{e^{\pi x}+e^{-\pi x}}\\ &= 2 \sum_{m=0}^{\infty} (-1)^m \left [\int_0^{\infty}dx \, e^{-[(2 m+1) \pi+i k] x} +\int_0^{\infty}dx \, e^{-[(2 m+1) \pi-i k] x} \right ] \\ &= 2 \sum_{m=0}^{\infty} (-1)^m \left [\frac{1}{(2 m+1) \pi-i k} + \frac{1}{(2 m+1) \pi+i k} \right ]\\ &= 4\pi \sum_{m=0}^{\infty} \frac{(-1)^m (2 m+1)}{(2 m+1)^2 \pi^2+k^2}\\ &= \frac{1}{2 \pi}\sum_{m=-\infty}^{\infty} \frac{(-1)^m (2 m+1)}{\left (m+\frac12\right)^2+\left(\frac{k}{2 \pi}\right)^2} \end{align}$$

By the residue theorem, the sum is equal to the negative sum of the residues at the non-integer poles of

$$\pi \csc{(\pi z)} \frac{1}{2 \pi}\frac{2 z+1}{\left ( z+\frac12\right)^2+\left (\frac{k}{2 \pi}\right)^2}$$

which are at $z_{\pm}=-\frac12 \pm i \frac{k}{2 \pi}$. The sum is therefore

$$-\frac12\csc{(\pi z_+)} - \frac12 \csc{(\pi z_-)} = -\Re{\left [\frac{1}{\sin{\pi \left (-\frac12+i \frac{k}{2 \pi}\right )}}\right ]} = \text{sech}{\left ( \frac{k}{2}\right)}$$

Solution 2:

Let's compute $\newcommand{\sech}{\mathrm{sech}}(-1)^n\int_{-\infty}^\infty\frac{\sech(n\pi x)}{x^2+1}\mathrm{d}x$ .

$\sech(n\pi x)$ has simple poles at $x=\frac{2k+1}{2n}i$, with residue $\frac{(-1)^k}{n\pi i}$ .

$\frac1{x^2+1}$ has simple poles at $x=\pm i$, with residue $\frac1{\pm2i}$ .

$\hspace{8mm}$

Note that the integrals along the vertical segments vanish at $\infty$.

The integral along the upper rectangular contour is the integral in question: $\int_{-\infty}^\infty\frac{\sech(n\pi x)}{x^2+1}\mathrm{d}x$, minus the Principal Value integral along the red path: $\color{#C00000}{\int_{-\infty+i}^{\infty+i}\frac{\sech(n\pi x)}{x^2+1}\mathrm{d}x}$, minus $\pi i$ times the residue at $x=i$: $\color{#00A000}{\pi i\times(-1)^{n}\frac1{2i}=(-1)^{n}\frac\pi2}$. This is also $2\pi i$ times the sum of the residues inside the contour: $\color{#0000FF}{\sum\limits_{k=0}^{n-1}\frac{(-1)^k}{n}\raise{6pt}{\frac2{1-\left(\frac{2k+1}{2n}\right)^2}}}$ .

The integral along the upper rectangular contour is the integral in question: $\int_{-\infty}^\infty\frac{\sech(n\pi x)}{x^2+1}\mathrm{d}x$, minus the Principal Value integral along the red path: $\color{#C00000}{\int_{-\infty-i}^{\infty-i}\frac{\sech(n\pi x)}{1+x^2}\mathrm{d}x}$, plus $\pi i$ times the residue at $x=-i$: $\color{#00A000}{\pi i\times-(-1)^{n}\frac1{2i}=-(-1)^{n}\frac\pi2}$. This is also $-2\pi i$ times the sum of the residues inside the contour: $\color{#0000FF}{\sum\limits_{k=0}^{n-1}\frac{(-1)^k}{n}\raise{6pt}{\frac2{1-\left(\frac{2k+1}{2n}\right)^2}}}$ .

Putting these together gives $$ \begin{align} &\int_{-\infty}^\infty\frac{\sech(n\pi x)}{x^2+1}\mathrm{d}x\\ &=\frac12\left( \color{#C00000}{(-1)^n\int_{-\infty}^\infty\frac{\sech(n\pi x)}{x^2+2ix}\mathrm{d}x} +\color{#00A000}{(-1)^n\frac\pi2} +\color{#0000FF}{\sum_{k=0}^{n-1}\frac{(-1)^k}{n}\frac2{1-\left(\frac{2k+1}{2n}\right)^2}} \right)\\ &+\frac12\left( \color{#C00000}{(-1)^n\int_{-\infty}^\infty\frac{\sech(n\pi x)}{x^2-2ix}\mathrm{d}x} +\color{#00A000}{(-1)^n\frac\pi2} +\color{#0000FF}{\sum_{k=0}^{n-1}\frac{(-1)^k}{n}\frac2{1-\left(\frac{2k+1}{2n}\right)^2}} \right)\\ &=(-1)^n\int_{-\infty}^\infty\frac{\sech(n\pi x)}{x^2+4}\mathrm{d}x +(-1)^n\frac\pi2 +8n\sum_{k=0}^{n-1}\frac{(-1)^k}{(2n)^2-(2k+1)^2}\\ &=\small(-1)^n\left(\frac12\int_{-\infty}^\infty\frac{\sech(2n\pi x)}{x^2+1}\mathrm{d}x +\frac\pi2\right) +2\sum_{k=0}^{n-1}(-1)^k\left(\frac1{2n+2k+1}+\frac1{2n-2k-1}\right)\\ &=\small(-1)^n\left(\frac12\int_{-\infty}^\infty\frac{\sech(2n\pi x)}{x^2+1}\mathrm{d}x +\frac\pi2 -4\sum_{k=0}^{n-1}(-1)^k\frac1{2k+1}+2\sum_{k=0}^{2n-1}(-1)^k\frac1{2k+1}\right)\tag{1} \end{align} $$ Rearranging $(1)$ yields $$ \begin{align} &(-1)^n\int_{-\infty}^\infty\frac{\sech(n\pi x)}{x^2+1}\mathrm{d}x +4\sum_{k=0}^{n-1}(-1)^k\frac1{2k+1} -\pi\\ &=\frac12\left( \int_{-\infty}^\infty\frac{\sech(2n\pi x)}{x^2+1}\mathrm{d}x +4\sum_{k=0}^{2n-1}(-1)^k\frac1{2k+1} -\pi \right)\\[6pt] &\to0\tag{2} \end{align} $$ Therefore, $$ (-1)^n\int_{-\infty}^\infty\frac{\sech(n\pi x)}{x^2+1}\mathrm{d}x =\pi-4\sum_{k=0}^{n-1}(-1)^k\frac1{2k+1}\tag{3} $$ Now, applying $(3)$, with $n=4$, to the question: $$ \begin{align} \int_0^\infty\frac{\mathrm{d}x}{\left(e^{\pi x}+e^{-\pi x}\right)\left(16+x^2\right)} &=\frac14\int_{-\infty}^\infty\frac{\mathrm{sech}(\pi x)\,\mathrm{d}x}{16+x^2}\\ &=\frac1{16}\int_{-\infty}^\infty\frac{\mathrm{sech}(4\pi x)\,\mathrm{d}x}{1+x^2}\\ &=\frac\pi{16}-\frac14\left(1-\frac13+\frac15-\frac17\right)\\ &=\frac\pi{16}-\frac{19}{105}\tag{4} \end{align} $$