Convergence of Improper Intergrals

Question

Find the value of the non-zero constant $c$ such that the following integral is convergent.

$\int_{-1}^\infty{\frac{e^{\frac{x}{c}}}{\sqrt{|x|}(x+2)}dx}$

I have no idea how to approach this, as I'm unable to integrate this function.

Answer 1

Let

$$I_1=\int_{-1}^0{\frac{e^{\frac{x}{c}}}{\sqrt{-x}(x+2)}dx}$$ and $$I_2=\int_{0}^\infty{\frac{e^{\frac{x}{c}}}{\sqrt{x}(x+2)}dx}$$ and the given integral is $I_1+I_2$ and since $$\frac{e^{\frac{x}{c}}}{\sqrt{-x}(x+2)}\sim_0\frac1{2\sqrt{-x}}$$ then $I_1$ exists, moreover we have $$\frac{e^{\frac{x}{c}}}{\sqrt{x}(x+2)}\sim_\infty \frac{e^{\frac{x}{c}}}{x^{3/2}}$$ so:

• if $c>0$ then $$\frac{e^{\frac{x}{c}}}{x^{3/2}}>1\quad\text{for $x$ large enough}$$ hence $I_2$ doesn't exist in this case,
• if $c<0$ then $$\frac{e^{\frac{x}{c}}}{x^{3/2}}=_\infty o\left(\frac{{1}}{x^{3/2}}\right)$$ and $I_2$ exists. Conclude.

Answer 2

When c<0, $\int_{1}^\infty{\frac{e^{\frac{x}{c}}}{\sqrt{|x|}(x+2)}dx}$ converges using Chartier's test: $\int_{-1}^\infty e^{\frac{x}{c}}$ converges (and therefore bounded), and ${\frac{1}{\sqrt{|x|}(x+2)}}$ is decreasing and tends to 0 as $x\to\infty$. When c>0, $e^{\frac{x}{c}}$ is much larger than ${{\sqrt{|x|}(x+2)}}$ and so, using comparison test, the integral diverges.

Then it remains to discuss convergence of $\int_{-1}^1{\frac{e^{\frac{x}{c}}}{\sqrt{|x|}(x+2)}dx}$. We see $\frac{e^{\frac{x}{c}}}{(x+2)}$ is bounded and positive, so we need only to discuss the convergence of $\int_{-1}^1{\frac{1}{\sqrt{|x|}}dx}$, and it equals $2\int_{0}^1{\frac{1}{\sqrt{x}}dx}$ which is convergent.