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The integral is:

$$f(a) = \int\limits_{-\infty}^\infty \frac{x e^{-a^2 x^2}}{\tanh(x)}dx$$

which seems to converge for all $a>0$. But I don't know how to get a sense of the function $f(a)$ such as writing it as a convergent series. The usual Taylor series has infinity for each term. Any ideas?

Edit:

I believe (using answers below) that when $a$ is close to zero we have (doing a substitution):

$$f(a) = \frac{1}{a^2}\int\limits_{-\infty}^\infty \frac{x e^{-x^2}}{\tanh (x/a)}dx$$

But using $\tanh(x/a)\rightarrow\operatorname{sign}(x)$ as $a\rightarrow 0^+$. So the above should become: $$f(a) \approx \frac{2}{a^2}\int\limits_{0}^\infty x e^{-x^2}dx = \frac{1}{a^2}$$

So that gives the behaviour of $f(a)$ when $a$ is small. But I don't know how to give extra terms. (Also not sure if this is mathematically correct). From numerical calculations I find that near $0$, $$ f(a)\approx \frac{1}{a^2} + \frac{\pi^6}{6} - \frac{\pi^4 }{60}a^2+\frac{\pi^6 }{252}a^4+... = \frac{1}{a^2}\sum\limits_{n=0}^\infty \frac{B_{2n} a^{2n}\pi^{2n}}{n!} $$ although apparently this doesn't converge?

Comment

The answers below are two asymptotic series depending on whether $a$ is small or large. These give good approximations if we truncate the summation before begins to diverge. In the mid-range, when $a^2=1/\pi$, these two sums become term-by-term equal and the closest the truncated sum get to the true answer of $f(1/\sqrt{\pi})$ is to about 1% error. Using both these sums, we can know any value to within about 1%-2% error, and if $a$ is small or large then much more accurately.

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    $\begingroup$ Just from calculations it seems like a good approximation when $a$ is close to zero is $f(a)\approx 1.6 + 1/a^2$. (Just putting values like $a=1/1000$ into Wolfram alpha. $\endgroup$
    – zooby
    May 24 at 22:34
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    $\begingroup$ Your conjectured result is correct, up to a missing $\pi^{2n} $ - I'll write an answer. $\endgroup$ May 25 at 3:31

2 Answers 2

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A Taylor series exists in powers of $1/a$: $$f(a) = \int\limits_{-\infty}^\infty \frac{x e^{-a^2 x^2}}{\tanh x}\,dx=a^{-2}\int\limits_{-\infty}^\infty x e^{-x^2}\,\text{cotanh}\, (x/a)\,dx$$ $$=\sum_{n=0}^{\infty}\frac{2^{2 n} B_{2 n} \,\Gamma \left(n+\frac{1}{2}\right)}{(2 n)!a^{2 n+1}},$$ with $B_{2n}$ the Bernoulli number.


To assess whether this asymptotic series is useful for $a\gtrsim 1$, below I plot the sum $\sum_{n=0}^{10}$, so the first eleven terms, for $a=1,2,3\ldots 10$ (blue data points), and compare with a numerical evaluation of the integral (blue curve).

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    $\begingroup$ Good stuff. Is it convergent? If range of convergence is $1/|a|<R$ for some $R$ it could mean that is doesn't converge near $a=0$. $\endgroup$
    – zooby
    May 24 at 21:25
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    $\begingroup$ Also in terms of series of $_1F_1$ hypergeometric functions $$f(a)=\sum_{n=0}^\infty\frac{2(1-2^{2n-1})\,B_{2n}}{(2n)!\,|a|^{2n+1}}\cdot\,_1F_1(n+1/2;1/2;1/(4a^2))$$ $\endgroup$ May 24 at 21:38
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    $\begingroup$ Alas the power series in powers of $1/a^2$ is only an asymptotic expansion at infinity; it does not converge for any $a$. [That is to be expected, because the series was obtained by integrating termwise the Taylor expansion of $\tanh x$ about $x=0$, and this Taylor expansion converges only for $|x| < \pi/2$ (the distance from $x=0$ to the nearest complex pole), whereas the integral is over all real $x$.] $\endgroup$ May 24 at 22:20
  • $\begingroup$ @JorgeZuniga Is that a convergent sum for all $a$? $\endgroup$
    – zooby
    May 26 at 17:09
  • $\begingroup$ I don't see how @JorgeZuniga 's series agrees with the integral; for example, for $a=5$ that sum gives 0.2006, while the correct value is 0.3568. $\endgroup$ May 26 at 17:24
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To obtain the series in $a$, separate off the leading term proportional to $1/a^2 $ and expand the Gaussian instead of the hyperbolic cotangent: \begin{eqnarray} f(a)&=& 2\int_{0}^{\infty } dx\, x \ e^{-a^2 x^2 } \left[ \coth x -1 +1 \right] \\ &=& 2\int_{0}^{\infty } dx\, x \ e^{-a^2 x^2 }+ 2\sum_{n=0}^{\infty } \frac{(-a^2)^{n}}{n!} \int_{0}^{\infty} dx\, x^{2n+1} [\coth x -1] \\ &=& \frac{1}{a^2 } + 2\sum_{n=0}^{\infty } \frac{(-a^2)^{n}}{n!} \frac{ (-1)^n \pi^{2n+2} B_{2n+2} }{2n+2} \\ &=& \frac{1}{a^2} + \frac{1}{a^2} \sum_{n=1}^{\infty } \frac{(\pi a)^{2n} B_{2n} }{n!} \\ &=& \frac{1}{a^2} \sum_{n=0}^{\infty } \frac{(\pi a)^{2n} B_{2n} }{n!} \end{eqnarray} Also this is only an asymptotic expansion - the Bernoulli numbers diverge more rapidly than $n!$ at large $n$.

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    $\begingroup$ Wow. I take it the coth x integral is just one you have to know. I take it also it's an asymptotic sum as someone else mentioned. ( I tried it out numerically and it seems to diverge for all x after long enough - although that could just be machine errors). $\endgroup$
    – zooby
    May 25 at 3:56
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    $\begingroup$ Well, I suppose there's no reason an integral should have a representation as a convergent series, I guess. $\endgroup$
    – zooby
    May 25 at 4:05
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    $\begingroup$ @zooby - Since $\coth x -1 = 1/(e^{2x} -1)$, the integral is one of the pretty standard ones that lead to Bernoulli polynomials ... $\endgroup$ May 25 at 4:06
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    $\begingroup$ By truncating the expansion of $e^{-a^2x^2}$ with a remainder term having known bound, you should be able to do that for the final expansion too. Then you will have rigorous bounds on $f(a)$. $\endgroup$ May 25 at 4:10

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