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 $v$.
Best Answer
We have that by $\log (1+x)=x+O(x^2)$ and $e^x=1+x+O(x^2)$
$$\left(1-\left(\frac{\lambda}{n}\right)^2\right)^{n-1}=e^{(n-1)\log\left(1-\left(\frac{\lambda}{n}\right)^2\right)}=e^{-\frac{(n-1)\lambda^2}{n^2}+O\left(\frac1{n^3}\right)}=1-\frac{(n-1)\lambda^2}{n^2}+O\left(\frac1{n^3}\right)$$
therefore
$$(n-1)\left[1-\left(1-\left(\frac{\lambda}{n}\right)^2\right)^{n-1}\right]=\frac{(n-1)^2\lambda^2}{n^2}+O\left(\frac1{n^2}\right) \to \lambda^2$$