Estimating $\int_{0}^\infty \frac{\exp(-a/(x^2+b))}{x^2+c}\cos(dx) dx$ I need a good estimate for the integral
$$I(a,b,c,d):=\int_{0}^\infty \frac{\exp(-a/(x^2+b))}{x^2+c}\cos(dx) dx.$$
Here $a,b,c,d>0$ and $b<c$. Of course it would be optimal to be able to compute the integral but having seen the comments to Solving or estimating $\int_{-\infty}^\infty \frac{\exp(-a/(x^2+b))}{x^2+c}dx$ I think, there is no hope. If you could make $I(a,b,c,d)$ explicit I guess
$$2\lim_{d \to 0}I(a,b,c,d)$$
would be a solution to the other question. But nevertheless, there might be a good estimate. The numerator is bounded by $1$. I know that since the integral is not positive all the time you can't do this:
$$\left| \int_{0}^\infty \frac{\exp(-a/(x^2+b))}{x^2+c}\cos(dx) dx \right| \le \left| \int_{0}^\infty \frac{\cos(dx)}{x^2+c} dx \right|.$$
This is very sad because I can compute the latter integral:
$$\int_{0}^\infty \frac{\cos(dx)}{x^2+c} dx = \frac{\pi}{2} \frac{\exp(-\sqrt{c}d)}{\sqrt{c}}.$$
For my purpose it would be sufficient to get such an estimate. So I'm asking you if a similar estimate is possible (maybe decorate my false estimate with a factor to make it work). For my estimate it is important that you don't lose the exponential dependency on $d$, so please don't simply estimate the cosine by $1$.
 A: Here is some similarity between your attempt and a complex approach. Substitute $z=dx$ and write $$I=\frac d2\Re\int_\Bbb R\frac{\exp(-A/(z^2+B))}{z^2+C}e^{iz}\,dz$$ where $A=ad^2$, $B=bd^2$ and $C=cd^2$. Take a semi-circular contour on the upper-half plane so that the line integral on $[-R,R]$ gives us $I$. The integral along the arc $Re^{it}$ $(R\to\infty;0\le t\le\pi)$ tends to zero by Jordan's lemma, so $$I=\frac d2\cdot2\pi\Re i(\operatorname{Res}(f,i\sqrt C)+\operatorname{Res}(f,i\sqrt B))$$ by Cauchy's theorem ($f$ denotes the integrand of $I$). We have \begin{align}\operatorname{Res}(f,i\sqrt C)&=\lim_{z\to i\sqrt C}\frac{e^{-A/(z^2+B)}}{z+i\sqrt C}e^{iz}=\frac{e^{A/(C-B)}}{2i\sqrt C}e^{-\sqrt C}=\frac{e^{a/(c-b)}}{2id\sqrt c}e^{-d\sqrt c}\\\operatorname{Res}(f,i\sqrt B)&=\operatorname{Res}\left(\frac{e^{i(z+i\sqrt B)-A/((z+i\sqrt B)^2+B)}}{(z+i\sqrt B)^2+C},0\right)=e^{-d\sqrt b}\operatorname{Res}\left(\frac{e^{iz}e^{-A/(z(z+2id\sqrt b))}}{(z+id\sqrt b)^2+cd^2},0\right)\end{align} so (note that as expected, the first term is the same as what you have computed) $$I=\frac{\pi e^{a/(c-b)-d\sqrt c}}{2\sqrt c}-\pi de^{-d\sqrt b}\Im\operatorname{Res}\left(\frac{e^{iz}e^{-A/(z(z+2id\sqrt b))}}{(z+id\sqrt b)^2+cd^2},0\right).$$
