Is it possible to find $$ \int_{0}^{K} \exp\left(-\,{c_{1} \over x} - {c_{2} \over x^{2}}\right) \,\mathrm{d}x $$ or at least a good approximation to it, where $c_{1}$ and $c_{2}$ are positive and $K$ is a very large number $?$.
I tried to consider the change of variables $-u = -c_{1}/x - c_{2}/x^{2}$ but to no avail.
Note: Approximation can include known functions such as Erf, or Gamma, if necessary.
On
As said, series expansion could be the solution. Writing $$\exp\left(-\frac{a}{x}-\frac{b}{x^2}\right)=1+\sum_{n=1}^p \frac {c_n}{n! \,x^n}=f(x)$$ the $c_n$'s would be $$\left( \begin{array}{cc} 1 & -a \\ 2 & a^2-2 b \\ 3 & -a^3+6 a b \\ 4 & a^4-12 b a^2+12 b^2 \\ 5 & -a^5+20 b a^3-60 b^2 a \\ 6 & a^6-30 b a^4+180 b^2 a^2-120 b^3 \\ 7 & -a^7+42 b a^5-420 b^2 a^3+840 b^3 a \\ 8 & a^8-56 b a^6+840 b^2 a^4-3360 b^3 a^2+1680 b^4 \\ 9 & -a^9+72 b a^7-1512 b^2 a^5+10080 b^3 a^3-15120 b^4 a \\ 10 & a^{10}-90 b a^8+2520 b^2 a^6-25200 b^3 a^4+75600 b^4 a^2-30240 b^5 \end{array} \right)$$ and $$\int\exp\left(-\frac{a}{x}-\frac{b}{x^2}\right)\,dx=1-a\log(x)-\sum_{n=2}^p \frac {c_n}{(n-1)\,n! \,x^{n-1}}+ C$$
For sure, there is a problem with a lower bound equal to $0$.
If you are interested in large values of $K$, an asymptotic expansion is possible. Using integration by parts on the equivalent integral
$$I(c_1,c_2;K)=\int_{1/K}^{\infty}\exp[-c_1 t-c_2 t^2]\frac{dt}{t^2}$$
one can isolate the divergent terms as follows
$$I=(K-c_1\log K)e^{-c_1/K-c_2/K^2}+2c_2\int_{1/K}^{\infty}dt\exp[-c_1 t-c_2 t^2]-c_1^2\int_{1/K}^{\infty}dt\log t\exp[-c_1 t-c_2 t^2]-2c_1c_2\int_{1/K}^{\infty}t\log t\exp[-c_1 t-c_2 t^2]dt$$
This is an exact statement so far and all the integrals on the RHS possess a finite limit as $K\to \infty$. The first of the three integrals is easily obtainable:
$$\int_{1/K}^\infty dt\exp[-c_1 t-c_2 t^2]=\frac{e^{c_1^2/4c_2}}{\sqrt{c_2}}\text{erfc}\left(\frac{\sqrt{c_2}}{K}+\frac{c_1}{2\sqrt{c_2}}\right)$$
As a matter of fact, Mathematica states that all of these integrals can be evaluated in terms of derivatives of hypergeometric functions with respect to one of their arguments in the limit $K\to\infty$. To wit,
$$8c_2^{3/2}\int_{0}^{\infty}t\log t\exp[-c_1 t-c_2 t^2]dt=\sqrt{\pi } c_1 e^{\frac{c_1^2}{4 c_2}} \left(_1F_1^{(1,0,0)}\left(0,\frac{3}{2},-\frac{c_1^2}{4 c_2}\right)+(\log c_2+\gamma ) \text{erfc}\left(\frac{c_1}{2 \sqrt{c_2}}\right)-2+\log (4)\right)-2 \sqrt{c_2} \left(- ~_{1}F_1^{(1,0,0)}\left(1,\frac{1}{2},\frac{c_1^2}{4 c_2}\right)+\log c_2+\gamma \right)$$
and
$$-4c_2\int_{0}^{\infty}\log t\exp[-c_1 t-c_2 t^2]dt=\sqrt{\pi }\sqrt{c_2} e^{\frac{c_1^2}{4 c_2}} \left(_1F_1^{(1,0,0)}\left(0,\frac{1}{2},-\frac{\text{c1}^2}{4 c_2}\right)+(\log c_2+\gamma) \text{erfc}\left(\frac{c_1}{2 \sqrt{c_2}}\right)+\log 4\right)+c_1 ~_{1}F_1^{(1,0,0)}\left(1,\frac{3}{2},\frac{c_1^2}{4 c_2}\right)$$
where $_1F_1^{(1,0,0)}(a,b,z)=\frac{\partial}{\partial a}~_1F_1(a,b,z)$. Assembling these terms one obtains the first few terms of the asymptotic series for the result $$I(c_1,c_2)\sim (K-c_1\log K)e^{-c_1/K-c_2/K^2}+2\sqrt{c_2}e^{c_1^2/4c_2}\text{erfc}\left(\frac{\sqrt{c_2}}{K}+\frac{c_1}{2\sqrt{c_2}}\right)-c_1^2\int_{0}^{\infty}dt\log t\exp[-c_1 t-c_2 t^2]-2c_1c_2\int_{0}^{\infty}t\log t\exp[-c_1 t-c_2 t^2]dt+\mathcal{O}\left(\frac{\log K}{K}\right)$$