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given this integral:
$$\int_{0}^{\infty} \frac{x}{e^{x^2}} dx $$

How can I prove that it's converge? I know how to calculate this integral, but, I want to know how to prove that this integral converge (without calculate the value of it's integral) [I thought maybe about test converge of integrals]..

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    $\begingroup$ Let $u=x^2.{}{}{}{}$ $\endgroup$ – Thomas Andrews Feb 18 '18 at 14:43
  • $\begingroup$ "I know how to calculate this integral", how? $\endgroup$ – Jack Feb 18 '18 at 14:46
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    $\begingroup$ You can certainly use that $e^{x^2}\geq x^{4}/2$ for any $x$, to get an upper bound that aslo converges. $\endgroup$ – Thomas Andrews Feb 18 '18 at 14:47
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For example let

$$\int_{0}^{\infty} \frac{x}{e^{x^2}} dx =\int_{0}^{1} \frac{x}{e^{x^2}} dx +\int_{1}^{\infty} \frac{x}{e^{x^2}} dx $$

the first one is a proper integral and the second one converges by limit comparison with

$$\int_{1}^{\infty} \frac1{x^2}dx$$

since

$$\frac{\frac{x}{e^{x^2}}}{\frac1{x^2}}\to 0$$

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Define $f(x)=e^{-x^2}$ then $\frac{x}{e^{x^2}}=-\frac12f'(x)$ and you should be able to compute $$ \int_0^\infty\frac{x}{e^{x^2}}~dx=\int_0^\infty-\frac12f'(x)~dx=\lim_{R\to\infty}\int_0^R -\frac12f'(x)~dx $$

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substitute $$t=x^2$$ then you will have $$dt=2xdx$$

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To show an integral converges, you can just evaluate it and show that it goes to a finite number.

You say in your answer that "I know how to calculate this integral, but, I want to know how to prove that this integral converge (without calculate the value of it's integral) [I thought maybe about test converge of integrals]"

You only use the tests when the integral is super bad and almost impossible to evaluate. Here this is not the case, and so you can easily substitute and solve for it.

As I'll skip the details as you are aware how to solve the integral, the solution becomes 0.5, and this is enough to show convergence.

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You can easily notice that $e^{x^2}> x^3 \quad \forall \ x \in \Bbb R$ so you have that $\displaystyle\frac{x}{e^{x^2}}<\frac{x}{x^3}=\frac{1}{x^2}$, hence:

$$\int_1^\infty\frac{x}{e^{x^2}}~dx < \int_1^\infty\frac{1}{x^2}~dx \quad \text{that converges}$$

Since $\displaystyle\int_0^\infty\frac{x}{e^{x^2}}~dx=\int_0^1\frac{x}{e^{x^2}}~dx+\int_1^\infty\frac{x}{e^{x^2}}~dx$ and the first one is a proper integral you have the result

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Consider to solve it:

$$I(x) = \int_0^{+\infty} x e^{-x^2}\ dx$$

$$I(x) = \int_0^{+\infty} -\frac{1}{2} \frac{d}{dx} e^{-x^2}\ dx = -\frac{1}{2}\left(e^{-x^2}\bigg|_{0}^{+\infty}\right) = \frac{1}{2}$$

Hence it converges.

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