I tried my best to solve the following definite integration.
Struggling a lot, ended with no luck. I know similar but a bit little simplified identity. 
The question is how to change this identity for my case which include two extra constants or how to solve this integral in first place.
You want to compute $$I=\int_0^X J_1( \alpha t)\,e^{-b t^2}\,dt$$
As you did, let $\alpha t=x$ to make $$I={\frac 1 \alpha}\int_0^{ \alpha X} J_1( x)\,e^{-\beta x^2}\,dx \qquad \text{with}\qquad \beta=\frac{b}{\alpha^2 }$$
Now, use the expansion $$ J_1( x)=\sum_{m=0}^\infty \frac{(-1)^m }{m! \,2^{2 m+1} \, \Gamma (m+2)}x^{2 m+1}$$ which makes that you face now the problem of $$K_m=\int x^{2 m+1}\,e^{-\beta x^2}\,dx=-\frac{ \Gamma \left(m+1, \beta x^2\right) }{2\beta ^{(m+1)}} \,$$ and $$L_m=\int_0^{ \alpha X} x^{2 m+1}\,e^{-\beta x^2}\,dx=\frac{\Gamma (m+1)-\Gamma \left(m+1,{ \beta \alpha ^2}X^2\right) }{2 \beta ^{m+1} }$$