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The following integral shows up when investigating spin glasses. I have been unable to solve it, other than numerically.

$$ \int_{-\infty}^{\infty} x\,e^{-a x^2} \tanh(b x)\,dx $$

I tried rewriting $\tanh(x)$ as a sum, and then switching the integral and summation. But then the resulting sum after performing the integral is divergent. I also can't figure out how to do this integral using contour methods.

bob
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Eric M.
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    It is unlikely there is a closed form for this integral. As a special case, see here. Your integral is related to what are known as Mordell integrals. – KStarGamer Nov 20 '23 at 23:19

1 Answers1

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$$I=\int_{-\infty}^{+\infty} x\,e^{-a x^2} \tanh(b x)\,dx$$

Let $bx=t$ and $c=\frac a {b^2}$ $$I=\frac 2 {b^2}\int_{0}^{\infty} t\, e^{-c t^2} \tanh (t)\,dt$$

Since, for $t>0$ $$\tanh(t)=1-\frac{2}{e^{2 t}+1}=1-2\sum_{n=0}^\infty (-1)^n\,e^{-2 (n+1) t}$$

and use $$J_n=\int_0^\infty t\, e^{-c t^2} \,e^{-2 (n+1) t}\,dt$$ $$J_n=\frac{1}{2 c}-\frac{\sqrt{\pi } }{2 c^{3/2}}\,(n+1)\,e^{\frac{(n+1)^2}{c}}\, \text{erfc}\left(\frac{n+1}{\sqrt{c}}\right)$$

For large $n$ $$J_n=\frac{1}{4 n^2}-\frac{1}{2 n^3}+O\left(\frac{1}{n^4}\right)$$

$$\Bigg|\frac {J_{n+1}-\frac 1 {2c}}{J_{n}-\frac 1 {2c}}\Bigg|=1+\frac{c}{n^3}-\frac{3 c}{2 n^4}+O\left(\frac{1}{n^5}\right)$$

Claude Leibovici
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