User:Nty/sandbox

If you compare $$cf(x)$$ with a density function of normal distribution $$\mathcal{N}(m, 1)$$ then its density function is equal to $$\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-m)}$$ and also we have that $$\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(x-m)^2} = 1$$ (by definition of a density function). By rearranging we obtain that $$\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x-m)} \ \mathrm{d} x = \sqrt{2\pi}$$ on the other hand we knew that $$\int_{-\infty}^{\infty}e^{-\frac{1}{2}(x-m)} =\frac{1}{c}$$ Therefore, $$c =\frac{1}{\sqrt{2\pi}}$$ and $$cf(x)$$ is just a density function of a normal distribution $$\mathcal{N}(m, 1)$$. In the question they ask for $$\int_{-\infty}^{\infty}c xe^{-\frac{1}{2}(x-m)}$$, since we know that $$cf(x)$$ is a density function then $$\int_{-\infty}^{\infty}c xe^{-\frac{1}{2}(x-m)}$$ is a first moment (expectation, mean) of this normal distribution and is equal to $$m$$.

Expectation $$\int_{-\infty}^{\infty} x c e^{-\frac{1}{2}(x-m)^2} \ \mathrm{d}x = \left| x-m = t \ | \  \mathrm{d} x = \mathrm{d} t  \ | \ x= t+m \right| = c\int_{-\infty}^{\infty}(t+m)e^{-\frac{1}{2}t^2}\mathrm{d} t$$ $$= c\underbrace{\int_{-\infty}^{\infty} te^{-\frac{1}{2}t^2}\mathrm{d} t}_{=:I_1} + cm  \underbrace{\int_{-\infty}^{\infty}e^{-\frac{1}{2}t^2}\mathrm{d} t}_{=:I_2}$$

$$I_2$$ you calculate only once when you find the density function of a normal distribution $$\mathcal{N}(0, 1)$$ so $$I_2 = \sqrt{2\pi} = \frac{1}{c}$$

$$I_1$$ is $$ 0 $$ and since $$te^{-\frac{1}{2}t^2}$$ is an odd function (that is it satisfies $$f(-t) = -f(t)$$) and every integral of an odd function on the symmetric interval with respect to $$0$$ is equal to $$0$$

Thus $$\int_{-\infty}^{\infty} x c e^{-\frac{1}{2}(x-m)^2} = cm \frac{1}{c}=m$$