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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?

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    $\begingroup$ Did you prove that the $\int_0^1$ converges only for $a < 1$? It doesn't seem like your argument does. If you don't prove that, then you can't conclude the integral is never convergent. Same for $\int_1^{\infty}$. $\endgroup$ – idok Jan 26 '18 at 13:34
  • $\begingroup$ And which other method could I use to see if $\int_0^1$ converges for other values of $a$? I can't think of any $\endgroup$ – John Keeper Jan 26 '18 at 13:56
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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 $\forall a<3$.

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    $\begingroup$ I believe is should be $\frac {2}{x^{a-2}}$. $\endgroup$ – idok Jan 26 '18 at 14:07
  • $\begingroup$ @idok yes of course, thanks I fix! $\endgroup$ – gimusi Jan 26 '18 at 14:11
  • $\begingroup$ I don't see why $\frac{e^{-x^2}-e^{-3x^2}}{x^a}\sim \frac{2}{x^{a-2}}$. Could you elaborate here? Thank you $\endgroup$ – John Keeper Jan 26 '18 at 14:26
  • $\begingroup$ @JohnKeeper It is simply by Taylor's expansion, I've added some details. You don't need strictly this step but it is very useful to find the term to use in the comparison test. $\endgroup$ – gimusi Jan 26 '18 at 14:30
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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 $\forall a<3$.

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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 $$ J(\alpha) = \color{red}{\frac{\pi\left(1-\sqrt{3^{\alpha-1}}\right)}{2\cos\left(\frac{\pi\alpha}{2}\right)\,\Gamma\left(\frac{\alpha+1}{2}\right)}}.$$

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