Convergence of $ \int_0^{\infty} \frac{e^{-x^2}-e^{-3x^2}}{x^a}$.

Question

I want to study the convergence of the improper integral $$ \int_0^{\infty} \frac{e^{-x^2}-e^{-3x^2}}{x^a}$$To do so I used the comparison test with $\frac{1}{x^a}$ separating $\int_0^{\infty}$ into $\int_0^{1} + \int_1^{\infty}$.

For the first part, $\int_0^{1}$, I did $$\lim_{x\to0} \frac{\frac{e^{-x^2}-e^{-3x^2}}{x^a}}{\frac{1}{x^a}}=0$$ Therefore $\int_0^{1}\frac{e^{-x^2}-e^{-3x^2}}{x^a}$ converges for $a<1$, since $\int_0^{1}\frac{1}{x^a}$ converges for $a<1$

For the second part I did the same, and got that $\int_1^{\infty}\frac{e^{-x^2}-e^{-3x^2}}{x^a}$ converges for $a>1$. This means that the initial improper integral does not converge for any $a$, is this correct?

Answer 1

For the first note that for $x\to0$

$$e^{-x^2}=1-x^2+o(x^2) \quad \quad e^{-3x^2}=1-3x^2+o(x^2)$$

$$\implies \frac{e^{-x^2}-e^{-3x^2}}{x^a}= \frac{2x^2+o(x^2)}{x^a}\sim \frac{2}{x^{a-2}}$$

thus $\int_0^{1}$ converges by comparison with $\frac{1}{x^{a-2}}$ for $a-2<1$ that is $a<3$.

For the second note that for $x\to +\infty$

$$\forall b \in \mathbb{R} \quad\frac{e^{-x^2}-e^{-3x^2}}{x^b}\to 0$$

thus $\int_1^{+\infty}$ converges by comparison with $\frac{1}{x^{2}}$ $\forall a$.

Therefore $\int_0^{\infty} \frac{e^{-x^2}-e^{-3x^2}}{x^a}$ converges $\forall a<3$.

Answer 2

It’s easy to see that your integral can be written as: $$I=\int_0^\infty e^{-x^2}x^{-a}dx-\int_0^\infty e^{-3x^2}x^{-a}dx$$ Then you compute the two parts of the integral separately: $$\int_0^\infty e^{-x^2}x^{-a}dx\overbrace{=}^{x^2=z}\frac 12\int_0^\infty e^{-z}z^{-\frac{a-1}2}dz=\frac 12\Gamma\left(\frac{1-a}2\right)\tag{1}$$ $$\int_0^\infty e^{-3x^2}x^{-a}dx\overbrace{=}^{3x^2=z}\frac{\sqrt{3^{a+1}}}6\int_0^\infty e^{-z}z^{-\frac {a-1}2}dz=\frac{\sqrt{3^{a+1}}}6\Gamma\left(\frac{1-a}2\right)\tag{2}$$ Hence your integral is simply $$I=\left(\frac 12-\frac{\sqrt{3^a+1}}6\right)\Gamma\left(\frac{1-a}2\right)$$ Where $\Gamma(\cdot)$ is the Euler Gamma Function.

And $I$ converges $\forall a<3$.

Answer 3

Your problem boils down to computing $$ I(\alpha)=\int_{0}^{+\infty}\frac{1-e^{-z^2}}{z^\alpha}\,dz = \frac{1}{2}\int_{0}^{+\infty}\frac{1-e^{-z}}{z^{\frac{\alpha+1}{2}}}\,dz. $$ In a right neighbourhood of the origin $\frac{1-e^{-z}}{z^{\frac{\alpha+1}{2}}}$ behaves like $z^{\frac{1-\alpha}{2}}$ and in a left neighbourhood of $+\infty$ it behaves like $z^{-\frac{1+\alpha}{2}}$, hence $I(\alpha)$ is convergent as soon as $\alpha\in(1,3)$, and in such a case $$I(\alpha)=-\frac{1}{2}\Gamma\left(\frac{1-\alpha}{2}\right)=-\frac{\pi}{2\cos\left(\frac{\pi\alpha}{2}\right)\,\Gamma\left(\frac{1+\alpha}{2}\right)}.$$ Similarly, the convergence of $$ J(\alpha)=\int_{0}^{+\infty}\frac{e^{-z^2}-e^{-3z^2}}{z^\alpha}\,dz $$ only depends on the integrability of the integrand function in a right neighborhood of the origin, and for any $\alpha < 3$ we have $$ J(\alpha) = \color{red}{\frac{\pi\left(1-\sqrt{3^{\alpha-1}}\right)}{2\cos\left(\frac{\pi\alpha}{2}\right)\,\Gamma\left(\frac{\alpha+1}{2}\right)}}.$$