$f(x)=\alpha e^{-x^2-\beta x}$ Find $\alpha$ and $\beta$ . Expectation is given.

Let the probability density function of a random variable $$X$$ be given by

$$f(x)=\alpha e^{-x^2-\beta x}\ \ \ \ \ \ \infty

If $$E(X)= -\dfrac{1}{2}$$ then

$$(A) \ \alpha =\dfrac{e^{\frac{-1}{4}}}{\sqrt \pi}$$ and $$\beta=1$$

$$(B) \ \alpha =\dfrac{e^{\frac{-1}{4}}}{\sqrt \pi}$$ and $$\beta=-1$$

$$(C) \ \alpha =e^\frac{-1}{4}\sqrt \pi$$ and $$\beta=-1$$

$$(D) \ \alpha =e^\frac{-1}{4}\sqrt \pi$$ and $$\beta=1$$

The way I did this question:

$$\int_{-\infty}^{\infty}f(x)dx=\int_{-\infty}^{\infty}\alpha e^{-(x^2+\beta x +\frac{\beta^2}{4}-\frac{\beta^2}{4})}dx=\int_{-\infty}^{\infty}\alpha e^{-(x+\frac{\beta}{2})^2}e^{\frac{\beta^2}{4}}dx=1$$

put $$y=(x+\frac{\beta}{2})$$

$$\int_{-\infty}^{\infty}\alpha e^{-(y)^2}e^{\frac{\beta^2}{4}}dx=\alpha\sqrt{\pi}e^{\frac{\beta^2}{4}}=1\implies \alpha=\dfrac{e^{\frac{-\beta^2}{4}}}{\sqrt{\pi}}$$

Now using $$E(X)=\int_{-\infty}^{\infty}x\alpha e^{-(x+\frac{\beta}{2})^2}e^{\frac{\beta^2}{4}}dx$$

put $$y=(x+\frac{\beta}{2})$$

$$E(X)=\int_{-\infty}^{\infty}(y-\frac{\beta}{2})\alpha e^{-(y)^2}e^{\frac{\beta^2}{4}}dx=\underbrace{\alpha e^{\frac{\beta^2}{4}} \int_{-\infty}^{\infty}(y) e^{-(y)^2}dx}_{\text {odd function}}-\frac{\beta}{2}\underbrace{\int_{-\infty}^{\infty}\alpha e^{-(y)^2}e^{\frac{-\beta^2}{4}}}_{\text{pdf}}dx$$

$$0-\dfrac{\beta}{2}=-\dfrac{1}{2} \implies \fbox {\beta =1}$$ and $$\fbox{\alpha=\dfrac{e^{\frac{-1}{4}}}{\sqrt{\pi}}}$$

Somehow after doing rigorous calculations, I got this answer but this question came in 2 marks so I am looking for a quick way out. Any alternate solution which less time consuming ?

$$f(x)$$ is in the form of a normal PDF: $$\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-(x-\mu)^2/2\sigma^2)$$, which should be equal to $$\alpha\exp(-x^2-\beta x)$$. Coefficients should match in LHS and RHS, so $$2\sigma^2=1$$ and $$E[X]=\mu=-1/2$$. Then, LHS becomes $$\frac{1}{\sqrt{\pi}}\exp(-x^2+2\mu x - \mu^2)=\alpha\exp(-x^2-\beta x)$$ which gives you $$\beta=1$$ and $$\alpha=\frac{\exp(-1/4)}{\sqrt{\pi}}$$.