Is $ \int_{-2}^{2}\frac{e^{-x}}{\sqrt{x+2}}\text dx $ convergent?

Question

Is $ \int_{-2}^{2}\frac{e^{-x}}{\sqrt{x+2}}\,\text dx $ convergent?

$$\int_{-2}^{2}\frac{e^{-x}}{\sqrt{x+2}}\,\text dx=\int_{-2}^{0}\frac{e^{-x}}{\sqrt{x+2}}\,\text dx+\int_{0}^{2}\frac{e^{-x}}{\sqrt{x+2}}\,\text dx$$

So the convergence depends of $\int_{-2}^{0}\frac{e^{-x}}{\sqrt{x+2}}\,\text dx$ :

$$\int_{-2}^{0}\frac{e^{-x}}{\sqrt{x+2}}\,\text dx=\lim_{a\to -2}\, \int_{a}^{0}\frac{e^{-x}}{\sqrt{x+2}}\,\text dx$$

Wolfram says the integral converges, so I triend to find $f(x)\geq\frac{e^{-x}}{\sqrt{x+2}}$ and easy to integrate.

My first idea was $\frac{1}{\sqrt{2+x}}$ but this function in $[-2,0]$ is not greater than $\frac{e^{-x}}{\sqrt{x+2}}$.

I'm stuck in this point, so can you help me or give me some hint? Thanks.

Answer 1

We have that

$$\int_{-2}^{2}\frac{e^{-x}}{\sqrt{x+2}}\text dx=\int_{0}^{4}\frac{e^{-{(y-2)}}}{\sqrt{y}}\text dy$$

and for $y\to 0^+$

$$\frac{e^{-{(y-2)}}}{\sqrt{y}}\sim \frac{e^{2}}{\sqrt{y}}$$

thus the given integral converges by limit comparison test with

$$\int_{0}^{4}\frac{1}{\sqrt{y}}\text dy$$

Answer 2

Your integral is $e^{-2}\int_0^4 x^{-1/2}e^{-x} \text dx\le e^{-2}\int_0^4 x^{-1/2}\text dx=2e^{-2}[\sqrt{x}]^4_0=4e^{-2}$, so is finite.