An integral equation with a singular kernel Solve the following integral equation for $h(t),~~ -1\le t\le 1$:
$$\int^1_{-1}\frac{h(t)}{\tanh[\lambda(x-t)]}dt=x, ~~~~-1\le x\le 1,$$
where the singular integral is understood in the sense of Cauchy principle value, and $\lambda>0$ is a constant. When $\lambda\to 0$, the above equation simplifies to
$$\int^1_{-1}\frac{h(t)}{\lambda(x-t)}dt=x, $$
whose solution is easily verified to be $h(t)=\frac{\lambda}{\pi} \sqrt{1-t^2}$. But away from this limit,
(1). Is it possible to get a simple analytic solution for $h(t)$, e.g. a power series in $\lambda$ with known coefficients, or an integral expression?
(2). Ultimately, I want to obtain a good lower bound on $h(0)$. By some intuition from the physical side, I seem to know that $h(0)$ is monotonously increasing in $\lambda$, and $h(0)\to1/2$ when $\lambda\to+\infty$. So a good lower bound should at least qualitatively capture those known features.
Any comments, ideas, or references are welcomed, thank you.
 A: Here's a partial approach:
Make the substitution $u = \lambda(x-t)$ to get
$$\int_{\lambda(x-1)}^{\lambda(x+1)}\dfrac{h\left(x - \frac u\lambda\right)}{\tanh u}\,du = x$$
Now assume that $h(t) = \sum_{n=0}^\infty h_nt^n$. Glossing over such silly details as whether or not exchange of the infinite sum and integral is actually justifiable, this gives
$$\sum_{n=0}^\infty h_n \int_{\lambda(x-1)}^{\lambda(x+1)}\dfrac{\left(x - \frac u\lambda\right)^n}{\tanh u}\,du = x$$
$$\sum_{n=0}^\infty h_n\sum_{k=0}^n{n\choose k}x^{n-k}\left(\frac{-1}\lambda\right)^k\int_{\lambda(x-1)}^{\lambda(x+1)}\frac {u^k}{\tanh u}\,du = x$$
Unfortunately $\frac {u^k}{\tanh u}$ does not appear to have a nice antiderivative. You can instead write $$\coth u = \frac 1u + \sum_{i=1}^\infty \dfrac{(-1)^{i-1}2^{2i}B_i}{(2i)!}x^{2i-1}$$
where the $B_i$ are the Bernoulli numbers.
Substituting in, performing the integration and rearranging the summations to write is a power series in $x$, one see that the coefficient of $x$ will be $1$ while all the other coefficients must be $0$. This gives a series of enquations in the $h_n$ which can be solved to give the solution you are after.
