Integral $\mathcal{P}\int_{-\infty}^{\infty}\frac{\tanh\left(\frac{1}{x^2}\right)}{x-k}\,dx$

Question

I need help solving the integral

$$\mathcal{I}(k)=\mathcal{P}\int_{-\infty}^{\infty}\frac{\tanh\left(\frac{1}{x^2}\right)}{x-k}\,dx$$

for $k>0$. The $\mathcal{P}$ denotes the Cauchy principal value. Mathematica has been unhelpful, but numerical tests show that it converges. I would also be happy with a series or asymptotic solution.

Edit: I attempt to solve for the asymptotic behaviour using Maxim's useful comment. Note that $$\mathcal{I}(k)\sim-\frac{1}{k}\int_{-\infty}^{\infty}\tanh\left(\frac{1}{x^2}\right)\,dx$$ for large $k$. Then $$\mathcal{I}(k)=-\frac{1}{k}\int_{-\infty}^{\infty}\tanh\left(\frac{1}{x^2}\right)\,dx+\int_{-\infty}^{\infty}\frac{\tanh\left(\frac{1}{x^2}\right)}{x-k}\,dx+\frac{1}{k}\int_{-\infty}^{\infty}\tanh\left(\frac{1}{x^2}\right)\,dx$$ $$\mathcal{I}(k)=-\frac{1}{k}\int_{-\infty}^{\infty}\tanh\left(\frac{1}{x^2}\right)\,dx+\int_{-\infty}^{\infty}\tanh\left(\frac{1}{x^2}\right)\left[\frac{1}{x-k}+\frac{1}{k}\right]\,dx$$ $$\mathcal{I}(k)=-\frac{1}{k}\int_{-\infty}^{\infty}\tanh\left(\frac{1}{x^2}\right)\,dx+\frac{1}{k}\int_{-\infty}^{\infty}\tanh\left(\frac{1}{x^2}\right)\left[\frac{x/k}{x/k-1}\right]\,dx$$ $$\mathcal{I}(k)=-\frac{1}{k}\int_{-\infty}^{\infty}\tanh\left(\frac{1}{x^2}\right)\,dx-\frac{1}{k^2}\int_{-\infty}^{\infty}x\tanh\left(\frac{1}{x^2}\right)\left[\frac{1}{1-x/k}\right]\,dx$$ $$\mathcal{I}(k)\sim-\frac{1}{k}\int_{-\infty}^{\infty}\tanh\left(\frac{1}{x^2}\right)\,dx-\frac{1}{k^2}\int_{-\infty}^{\infty}x\tanh\left(\frac{1}{x^2}\right)\sum_{j=0}^{\infty}\left(\frac{x}{k}\right)^j\,dx$$ $$\mathcal{I}(k)\sim-\sum_{j=0}^{\infty}\frac{\int_{-\infty}^{\infty}x^j\tanh\left(\frac{1}{x^2}\right)}{k^{j+1}}$$ $$\mathcal{I}(k)\sim-\sum_{j=0}^{\infty}\frac{\int_{-\infty}^{\infty}x^{2j}\tanh\left(\frac{1}{x^2}\right)}{k^{2j+1}}$$ $$\mathcal{I}(k)\sim-2\sum_{j=0}^{\infty}\frac{\int_{0}^{\infty}x^{2j}\tanh\left(\frac{1}{x^2}\right)}{k^{2j+1}}$$ Edit: This still doesn't work. My integrals diverge for $j>0$. Perhaps useful: $$\tanh(x)=\sum_{j=0}^{\infty}2\frac{(-1)^{j}}{\pi^{2j+2}}\left(4^{j+1}-1\right)\zeta(2j+2)x^{2j+1}$$

Answer 1

Here is a different approach, without analytical continuation.

\begin{gather} I=\mathcal{P}\int_{-\infty}^\infty\frac{\tanh\frac{1}{x^2}}{x-k}\,dx=-\frac{1}{k}\mathcal{P}\int_0^\infty\frac{\tanh t\,dt}{\sqrt{t}(t-k^{-2})}=-\frac{1}{k}\mathcal{P}\int_0^\infty\frac{dt}{\sqrt{t}(t-k^{-2})}+\frac{2}{k}\mathcal{P}\int_0^\infty\frac{dt}{\sqrt{t}(e^{2t}+1)(t-k^{-2})}=I_1+I_2 \end{gather} Now, the contour your are going to use to take those two integrals is depicted in Fig.1

You can easily check that the integral along the small circle round point $t=k^{-2}$ vanishes (due to regular branch of $\sqrt{t}$ sign change). This way you immediately see that $I_1=0$. Now you compute $I_2$. This is a typical $\zeta$-kernel type integral. Denote: \begin{gather} f_2(t)=\frac{1}{\sqrt{t}(e^{2t}+1)(t-k^{-2})},\quad \oint f_2(t)\,dt=2I_2=2\pi i\sum\limits_{n=-\infty}^\infty \underset{t=\frac{i\pi}{2}(2n+1)}{\rm res}f_2(t) \end{gather} Next, \begin{gather} \underset{t=\frac{i\pi}{2}(2n+1)}{{\rm res}}f_2(t)=\sqrt{\frac{2}{\pi^3}}\frac{1}{2n+1+2ik^{-2}/\pi}(|2n+1|)^{-1/2} \begin{cases} e^{i\pi/4},\ \ n\geq0,\\ e^{-i\pi /4}, n<0. \end{cases} \end{gather} And the whole thing is done: \begin{gather} \sum_{n=0}^\infty\underset{t=\frac{i\pi}{2}(2n+1)}{{\rm res}}f_2(t)=e^{i\pi/4}\sqrt{\frac{2}{\pi^3}}\sum_{n=0}^\infty\frac{1}{2n+1+2ik^{-2}/\pi}(2n+1)^{-1/2}\\ =e^{i\pi/4}\sqrt{\frac{2}{\pi^3}}\sum_{l=0}^\infty(-1)^l\Big(\frac{2i}{k^2\pi}\Big)^l\sum\limits_{n=0}^\infty\frac{1}{(2n+1)^{l+3/2}}= e^{i\pi/4}\sqrt{\frac{2}{\pi^3}}\sum_{l=0}^\infty(-1)^l\big[1-2^{l+3/2}\big]\Big(\frac{2i}{k^2\pi}\Big)^l\zeta(l+3/2) \end{gather} In the same manner you compute the sum over the lower poles: \begin{gather} \sum_{{\rm lower\ poles}}= -\left(\sum_{{\rm upper\ poles}}\right)^*. \end{gather} Hence: \begin{gather} 2\pi i\sum_{{\rm all\ poles}}= 4\pi\sqrt{\frac{2}{\pi^3}}\sum_{l=0}^\infty \sin\left(\frac{\pi l}{2}-\frac{\pi}{4}\right)\frac{2^l}{\pi^l k^{2l}}\zeta(l-3/2). \end{gather} And the initial integral reads: \begin{gather} I=I_2 = 4\sqrt{\frac{2}{\pi}}\sum_{l=0}^\infty \sin\left(\frac{\pi l}{2}-\frac{\pi}{4}\right)\frac{2^l}{\pi^l k^{2l+1}}\zeta(l+3/2)\big[1-2^{l-3/2}\big]. \end{gather}

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Answer 2

The full series reads as follows: $$ \mathcal{I}(k)=\sum_{n=0}^{\infty}2^{n+1} \left(\sqrt{2}-2^{n+2}\right) \zeta \left(-n-\frac{1}{2}\right) \Gamma \left(-n-\frac{1}{2}\right)k^{-2n-1} $$ It is easy to check it numerically. If you are happy with this kind of result I can add more details later.