# How would you work out this integral as a series?

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.

• 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. May 24 at 22:34
• Your conjectured result is correct, up to a missing $\pi^{2n}$ - I'll write an answer. May 25 at 3:31

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). • 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 $a=0$. May 24 at 21:25
• Also in terms of series of $_1F_1$ hypergeometric functions $$f(a)=\sum_{n=0}^\infty\frac{2(1-2^{2n-1})\,B_{2n}}{(2n)!\,|a|^{2n+1}}\cdot\,_1F_1(n+1/2;1/2;1/(4a^2))$$ May 24 at 21:38
• 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 $x$.] May 24 at 22:20
• @JorgeZuniga Is that a convergent sum for all $a$? May 26 at 17:09
• 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. May 26 at 17:24

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 $$n$$.

• 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). May 25 at 3:56
• Well, I suppose there's no reason an integral should have a representation as a convergent series, I guess. May 25 at 4:05
• @zooby - Since $\coth x -1 = 1/(e^{2x} -1)$, the integral is one of the pretty standard ones that lead to Bernoulli polynomials ... May 25 at 4:06
• 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 $f(a)$. May 25 at 4:10