If $\displaystyle A=\int^{x}_{0}e^{zx}e^{-z^2}dz$ and If $\displaystyle B=\int^{x}_{0}e^{-\frac{z^2}{4}}dz$ Then relation between $A$ and $B$ is
Try: assuming $$f(x)=\int^{x}_{0}e^{zx}e^{-z^2}dx-\int^{x}_{0}e^{-\frac{z^2}{4}}dx$$
$$f'(x)=e^{-\frac{z^2}{2}}-e^{-\frac{z^2}{4}}$$
Could some help me to explain that my trial is right or not.
Also explain me correct solution. Thanks
Note that $$A\left(x\right)=\int_{0}^{x}\exp\left(-z^{2}+zx\right)dz=e^{x^{2}/4}\int_{0}^{x}\exp\left(-z^{2}+zx-x^{2}/4\right)dz$$ $$=e^{x^{2}/4}\int_{0}^{x}\exp\left(-\left(z-x/2\right)^{2}\right)dx\stackrel{z-x/2=u/2}{=}\frac{e^{x^{2}/4}}{2}\int_{-x}^{x}\exp\left(-u^{2}/4\right)du$$ $$=e^{x^{2}/4}\int_{0}^{x}\exp\left(-u^{2}/4\right)du=e^{x^{2}/4}B\left(x\right).$$