Finding relation between $A$ and $B$ is

Question

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

the integral $A$ is given by $${{\rm erf}\left(1/2,x\right)}\sqrt {\pi }{{\rm e}^{1/4,{x}^{2}}}$$ — Dr. Sonnhard Graubner, Feb 05 '18 at 13:36

score 3 · Accepted Answer · answered Feb 05 '18 at 14:01

Note that $$A\left(x\right)=\int_{0}^{x}\exp\left(-z^{2}+zx\right)dz=e^{x^{2}/4}\int_{0}^{x}\exp\left(-z^{2}+zx-x^{2}/4\right)dz$$ $$=e^{x^{2}/4}\int_{0}^{x}\exp\left(-\left(z-x/2\right)^{2}\right)dx\stackrel{z-x/2=u/2}{=}\frac{e^{x^{2}/4}}{2}\int_{-x}^{x}\exp\left(-u^{2}/4\right)du$$ $$=e^{x^{2}/4}\int_{0}^{x}\exp\left(-u^{2}/4\right)du=e^{x^{2}/4}B\left(x\right).$$

Finding relation between $A$ and $B$ is

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