Evaluation of a tricky integral involving the pdf of a normal distribution

Question

I tried to evaluate the following integral: $$-\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}e^{-aw}e^{-\frac{(w-\mu)^2}{2\sigma^2}}dw.$$ It seems that integration by parts does not work. Any help will be appreciated.

The following answer is given by my instructor: $$\mu-\frac{a}{2}\sigma^2$$

Answer 1

Completing the square on the other way: \begin{align*} aw+\frac{(w-\mu)^{2}}{2\sigma^{2}} &= \frac{w^{2}-2(\mu+a\sigma^{2})w+\mu^{2}}{2\sigma^{2}} \\ &= \frac{w^{2}-2(\mu+a\sigma^{2})w+(\mu+a\sigma)^{2}- 2a\mu \sigma^{2}-a^{2}\sigma^{4}}{2\sigma^{2}} \\ &= \frac{(w-\mu-a\sigma^{2})^{2}}{2\sigma^{2}}- a\left( \mu+\frac{a\sigma^{2}}{2} \right) \\ \end{align*}

Now \begin{eqnarray*} && -\frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp \left[ -aw-\frac{(w-\mu)^{2}}{2\sigma^{2}} \right] dw \\ &=& -\exp \left[ -a\left( \mu+\frac{a\sigma^{2}}{2} \right) \right] \times \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \exp \left[ -\frac{(w-\mu-a\sigma^{2})^{2}}{2\sigma^{2}} \right] dw\\ &=& -\sigma \, \exp \left[ -a\left( \mu+\frac{a\sigma^{2}}{2} \right) \right] \times \frac{1}{\sqrt{2\pi}\, \sigma} \int_{-\infty}^{\infty} \exp \left[ -\frac{x^{2}}{2\sigma^{2}} \right] dx\\ &=& -\sigma \, \exp \left[ -a\left( \mu+\frac{a\sigma^{2}}{2} \right) \right] \end{eqnarray*}

Answer 2

Let $x=\frac{w-\mu}{\sqrt{2}\sigma}$

$$-\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\exp\left[-\frac{(w-\mu)^2}{2\sigma^2}-aw\right]dw$$

$$=-\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}\exp\left[-x^2-a(\sqrt{2}\sigma x+\mu)\right]\sqrt{2}\sigma dx$$

$$=-\frac{\sigma}{\sqrt{\pi}}\exp\left[-a\mu\right]\int_{-\infty}^{\infty}\exp\left[-x^2-\sqrt{2}a\sigma x\right] dx$$

$$=-\frac{\sigma}{\sqrt{\pi}}\exp\left[-a\mu\right]\int_{-\infty}^{\infty}\exp\left[-\bigg(x+\frac{a\sigma}{\sqrt{2}}\bigg)^2+\frac{a^2\sigma^2}{2}\right] dx$$

$$=-\frac{\sigma}{\sqrt{\pi}}\exp\left[-a\mu+\frac{a^2\sigma^2}{2}\right]\int_{-\infty}^{\infty}\exp\left[-\bigg(x+\frac{a\sigma}{\sqrt{2}}\bigg)^2\right] dx$$

$$=-\frac{\sigma}{\sqrt{\pi}}\exp\left[-a\mu+\frac{a^2\sigma^2}{2}\right]\int_{-\infty}^{\infty}\exp\left[-x^2\right] dx$$

$$=-\frac{\sigma}{\sqrt{\pi}}\exp\left[-a\mu+\frac{a^2\sigma^2}{2}\right]\sqrt{\pi}$$

$$=-\sigma\exp\left[-a\mu+\frac{a^2\sigma^2}{2}\right]$$