Evaluating
$$
F_\nu(t) :=
\int_0^\infty \frac{u}{2\nu^2} \left( u+ \left( u^2+\nu^2 \right)^\frac{1}{2} \right)
\left(
e^{-u} – e^{-\left(u^2+\nu^2\right)^\frac{1}{2}}
\right) \sin \left( ut \right) \, \mathrm{d} u \, ,
$$
where $\nu\ge 0$ is a parameter.
Considering the simplistic limit when $\nu \to 0$, it can readily be shown that
$$
\lim_{\nu\to 0}
F_\nu (t) = \frac{1}{2} \int_0^\infty u\, e^{-u} \sin \left(ut\right) \, \mathrm{d} u
= \frac{t}{\left( 1+t^2 \right)^2} \, .
$$
This integral arises from solving a fluid mechanical problem involving dual integral equations. The full problem has previously been asked in a separate question.
What I tried is to express the sine and exponential functions in terms of series expansions.
This yield double sums involving hypergeometric functions.
However, an analytical evaluation of the resulting sums does not seen to be within reach.
Assuming that $\alpha \ll 1$ and performing Taylor expansion of $F_\nu(t)$ around $\alpha=0$ leads to a series of terms whose integrals diverge (except the terms $\propto \alpha^2$), making a Taylor-expansion-based approach obsolete.
I am wondering whether there is a clever way to handle this improper integral. Any help is highly appreciated.
Best Answer
One can make some progress in this problem by noticing that, after expanding the products in the integrand, all resulting individual parts of the sum can be expressed in terms of the more general integral
$$I(x,y)=\int_{0}^{\infty}e^{-xu}e^{-y\sqrt{u^2+v^2}}du$$
In fact, more explicitly one can write
$$\int_0^\infty u\sqrt{u^2+v^2}e^{-u}\sin{ut}~du= \text{Im}\frac{\partial^2I}{\partial x\partial y}\Bigg|_{(x,y)=(1+it,0)}$$ $$\int_0^\infty u^2e^{-\sqrt{u^2+v^2}}\sin{ut}~du= \text{Im}\frac{\partial^2 I}{\partial x^2}\Bigg|_{(x,y)=(it,1)}$$ $$\int_0^\infty u\sqrt{u^2+v^2}e^{-\sqrt{u^2+v^2}}\sin{ut}~du= \text{Im}\frac{\partial^2I}{\partial x\partial y}\Bigg|_{(x,y)=(it,1)}$$
We can hence focus on the evaluation of $I$. First note that $I$ can be expanded in a generating function-like form:
$$I(x,y)=\sum_{n=0}^{\infty}\frac{(-1)^n x^n}{n!}J_n(y)~~,~~ J_n(y):=\int_0^{\infty}u^n e^{-y\sqrt{u^2+v^2}}du$$
It is simple to show that the $J_n$'s obey the following recurrence relation
$$J_{n+2}(y)=\left(\frac{\partial^2}{\partial y^2}-v^2\right)J_n(y)$$
and therefore knowledge of $J_0, J_1$ allows determination of all the other moments. We can now write down the formal expression
$$I(x,y)=\sum_{n=0}^\infty x^{2n}\frac{\left(\frac{\partial^2}{\partial y^2}-v^2\right)^n}{(2n)!}J_0(y)-\sum_{n=0}^\infty x^{2n+1}\frac{\left(\frac{\partial^2}{\partial y^2}-v^2\right)^n}{(2n+1)!}J_1(y)$$
The first two moments are reasonably easy to evaluate, and in fact the first order one is elementary:
$$J_0(y)=vK_1(yv)~~,~~J_1(y)=\frac{1+vy}{y^2}e^{-vy}:= \sqrt{\frac{2}{\pi}} \frac{v^2}{\sqrt{vy}}K_{3/2}(vy)$$
We note here that it is possible to express $(d^2/dt^2-1)^n K_1(t)$ as a linear combination of $K_1(t), K_1'(t)$ due to the 2nd order ODE that Bessel-K functions satisfy:
$$\left(d^2/dt^2-1\right)^n K_1(t)=P_n(t)K_1(t)+Q_n(t)K_1'(t)$$
This could be useful since these functions can be reduced to polynomials, and the resulting infinite series are just polynomial generating functions. However, quite nontrivially and compactly, we can rewrite this in terms of higher order Bessel-K functions:
$$\boxed{\left(\frac{d^2}{dt^2}-1\right)^nK_1(t)=(2n-1)!!\frac{K_{n+1}(t)}{t^n}},$$ with $n!!$ denoting the double factorial, 3!! = 3, 5!! = 15, etc...
Similarly, we can rewrite the derivatives of $K_{3/2}(t)/\sqrt{t}$ as
$$\boxed{\left(\frac{d^2}{dt^2}-1\right)^n\frac{K_{3/2}(t)}{\sqrt{t}}=2^n n!\frac{K_{n+3/2}(t)}{t^{n+1/2}}}$$
Putting everything together:
$$\boxed{I(x,y)=v\sum_{n=0}^\infty\frac{(x^2v/y)^n}{(2n)!} (2n-1)!! K_{n+1}(vy)- \sqrt{\frac{2}{\pi}}v\sum_{n=0}^\infty\frac{(x^2v/y)^{n+1/2}}{(2n+1)!}2^n n!K_{n+3/2}(vy)}$$
To my knowledge, there is no available analytic formula for exponential generating functions of modified Bessel functions of the second kind, but the above connection is already pretty nice. Note that it's a bit more neat than expanding the exponential and sine functions separately and leads to just a single infinite sum of Bessel's (which also happen to be low level hypergeometrics).
Note: the two series have only been formally separated, they don't converge individually (by virtue of their individual integral representations not converging). To get a convergent answer, both series need to be taken into account.
The series has successfully been checked and found to to provide correct results when compared with the numerical integration of $I(x,y)$ for arbitrary $x$, $y$, and $v$.