Solve exponenital integral equation $$\frac{1}{\sqrt{2\pi}\sigma_1 }\int_x^\infty\exp(-\frac {t_1^2-1}{2\sigma_1^2})dt_1 + \frac{1}{\sqrt{2\pi}\sigma_2 }\int_x^\infty\exp(-\frac {t_2^2-1}{2\sigma_2^2})dt_2 = a $$
$$\sigma_1 , \sigma_2 \gt 0$$ 
Is there a way to solve for $x$ !?
please help
 A: As said in comments and answers, there is no explicit solution to the equation which, after integration write $$f(x)=\frac{1}{2} e^{\frac{1}{2 \sigma_1^2}} \text{erfc}\left(\frac{x}{\sqrt{2} \sigma
   _1}\right)+\frac{1}{2} e^{\frac{1}{2 \sigma_2^2}}
   \text{erfc}\left(\frac{x}{\sqrt{2} \sigma_2}\right)-a=0$$ $$f'(x)=-\frac{e^{\frac{1-x^2}{2 \sigma_2^2}}}{\sqrt{2 \pi } \sigma
   _2}-\frac{e^{\frac{1-x^2}{2 \sigma_1^2}}}{\sqrt{2 \pi }
   \sigma_1}$$ In the case where $x$ would be small, you could use as a very first approximation $$\text{erfc}(y)=1-\frac{2 y}{\sqrt{\pi }}+O\left(y^2\right)$$ and solve one equation for one unknown.
Otherwise, Newton method would be the simplest way to solve the equation.
A: Let $X_i\sim\mathcal N(1,\sigma_i^2)$ be independent for $i=1,2$, then the expression above is
$$\mathbb P(X_1>x) + \mathbb P(X_2>x) = a. $$
After standardizing we find that
$$\Phi\left(\frac{x-1}{\sigma_1}\right) + \Phi\left(\frac{x-1}{\sigma_2}\right)=a, $$
where $\Phi$ is the distribution function of the normal distribution with $\mu=0$, $\sigma^2=1$. In terms of the error function, this is
$$\operatorname{erf}\left(\frac{x-1}{\sqrt 2\sigma_1}\right)+\operatorname{erf}\left(\frac{x-1}{\sqrt 2\sigma_2}\right)=2(1-a). $$
This equation cannot be solved analytically, but can be approximated numerically in many ways.
