integrate $\int_{-\infty}^\infty e^{x-\frac{1}{2}\frac{(x+0.95)^2}{0.35^2}}dx $

Question

How can we integrate $$\int_{-\infty}^\infty e^{x-\frac{1}{2}\frac{(x+0.95)^2}{0.35^2}}dx ~~?$$Is it integrable function over $\mathbb R$?, Can we use the fact $\int_{-\infty}^\infty e^{ax-x^2}dx = \sqrt{\pi}e^{\frac{a^2}{4}}$

Answer 1

You can use the similar fact that

(i) $\int_{-\infty}^\infty e^{ax-bx^2}dx = \frac{\sqrt{\pi}}{\sqrt{b}}e^{\frac{a^2}{4b}}$

The expression in the exponent is:

$x-\frac{1}{2}\frac{(x+0.95)^2}{0.35^2} = x-\frac{x^2+1.9x + 0.95^2}{2\cdot (0.35)^2} = (1 + \frac{1.9x}{2\cdot (0.35)^2})x -\frac{1}{2\cdot (0.35)^2}x^2 + \frac{0.95^2}{2\cdot (0.35)^2}$

You can take

$e^{\frac{0.95^2}{2\cdot (0.35)^2}}$

Out of the integral to get

$\int_{-\infty}^\infty e^{x-\frac{1}{2}\frac{(x+0.95)^2}{0.35^2}}dx = e^{\frac{0.95^2}{2\cdot (0.35)^2}} \int_{-\infty}^\infty e^{(1 + \frac{1.9x}{2\cdot (0.35)^2})x -\frac{1}{2\cdot (0.35)^2}x^2}dx$

And apply (i) from here.