# 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 $$b. Of course it would be optimal to be able to compute the integral but having seen the comments to Solving or estimating $\int_{-\infty}^\infty \frac{\exp(-a/(x^2+b))}{x^2+c}dx$ I think, there is no hope. If you could make $$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$$.

• $$\left| \int_{0}^\infty \frac{\cos(dx)}{x^2+c} \mathrm dx \right|\le \int_{0}^\infty \frac{\left|\cos(dx)\right|}{x^2+c} \mathrm d x = \int_{0}^\infty \frac{1}{x^2+c} \mathrm dx = \frac{\pi}{2\sqrt c}.$$ Right? Commented Sep 29, 2021 at 9:40
• @SewerKeeper OP specifically asked not to use the estimate $|\cos(x)| \leq 1$ Commented Sep 29, 2021 at 9:42
• @SewerKeeper You need a $\sqrt{c}$ in the denominator and your "$=$" should be a "$\le$", but have you read my last sentence? Commented Sep 29, 2021 at 9:42
• Ops, didn't read the last line. Sorry ^^' Commented Sep 29, 2021 at 9:43
• might want a different letter to $d$ since it currently reads $\cos(dx)\,dx$ Commented Sep 29, 2021 at 9:48

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) $$I=\frac{\pi e^{a/(c-b)-d\sqrt c}}{2\sqrt c}-\pi de^{-d\sqrt b}\Im\operatorname{Res}\left(\frac{e^{iz}e^{-A/(z(z+2id\sqrt b))}}{(z+id\sqrt b)^2+cd^2},0\right).$$
• You say "The line integral on $[−R,R]$ tends to zero by Jordan's lemma". Don't you mean the integral over the arc? Commented Sep 29, 2021 at 11:19
• @principal-ideal-domain Yeah thanks, I wrote it the wrong way round. Theoretically, we can compute $I$ by computing the residue at zero, but that means we need to evaluate a very horrible-looking sum (to extract the $1/z$ coefficient, which is why I haven't expanded on that part). So a double/triple sum is definitely available but I'm not expecting anything more elegant. Commented Sep 29, 2021 at 11:41
• But wouldn't that give an answer to math.stackexchange.com/questions/4261677/… after taking the limit $d \to 0$? Commented Sep 29, 2021 at 11:53