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I have to find the integral of
$$int_M_0^infty q(m, mu, sigma) beta e^-beta(m-M_0),mathrmdm,$$
where $q(m, mu, sigma)$ is the normal cumulative distribution function, $M_0$ is a constant, $m$ is the variable, and $beta$, $mu$, and $sigma$ are parameters. I have done the integration using the error function as follows:

beginalign
int_M_0^infty q(m, mu, sigma) beta e^-beta(m-M_0),mathrmdm &=beta e^beta M_0 int_M_0^infty frac12 Bigg[ 1+operatornameerfBigg (frac(m-mu)sigma sqrt(2) Bigg ) Bigg] e^-beta m ,mathrmdm \
&=frac12 + frac12 beta e^beta M_0 int_M_0^infty operatornameerf Bigg (frac(m-mu)sigma sqrt(2) Bigg)e^- beta m,mathrmdm
endalign

Here I get stuck. Could anyone please help solving this?

I will present to you the solution in a sort of general step-by-step method. I hope you find it useful.

First of all, you have to use Partial Integration to get rid of the normal CDF function in the integral. Can you give it a go? If you have some difficulties, just ask me.

If you have done this correctly, you will end up with an integral of the form
$$ int_a^b e^a x^2 + b x + c,mathrmd x $$
which you can't solve or can you? It seems difficult but it turns out that this is actually quite easy if you allow the normal CDF in your answer. The main idea in solving this integral is the same as splitting the square. So, this integral can be simplified to
$$ int_a^b e^k(x + l)^2 + m,mathrmd x $$
and this is then solved as
$$ int_a^b e^k(x + l)^2 + m,mathrmd x = e^m int_a^b e^frac(x + l)^22(1/sqrt2 k)^2,mathrmd x = e^m cdot left[Phi(b) - Phi(a)right]cdot sqrt2pi (1/sqrt2k)^2 $$
where $Phi(z)$ is the CDF of a normal distribution with mean $-l$ and variance $1/(2k)$. This seems a bit random but all we did was actually writing the integrand to something which is similar to a normal pdf. Finally, note that we cannot simplify this any further in general. However, in your case, you have $b=infty$ and, thus, $Phi(b)=1$.

the integral is:

$I= int_M_0^infty q(m,mu,sigma). beta e^-beta (m-M_0) dm \
=frac12+frac12 beta e^beta M_0 * I_11$

where,
$I_11= int_M_0^infty erf(frac(m-mu)sigma sqrt2) e^- beta m dm\
=erf(frac(m-mu)sigma sqrt2) int_M_0^infty e^- beta m dm - int_M_0^infty fracddm erf(frac(m-mu)sigma sqrt2) int_M_0^infty e^- beta m dm dm \
=frac1beta erf(frac(m-mu)sigma sqrt2) e^- beta M_0 + frac1beta frac2sqrtpi int_M_0^infty e^frac(m-mu)^22 sigma^2 e^- beta m dm \
=frac1beta erf(frac(m-mu)sigma sqrt2) e^- beta M_0+ frac1beta frac2sqrtpi I_22$

where,
$I_22= int_M_0^infty e^frac(m-mu)^22 sigma^2 e^- beta m dm\
= int_M_0^infty e^fracm-(mu-beta sigma^2) ^22 sigma^2+ fracbeta2 sigma^2-2mu dm$