$\int_{-\infty}^{\infty}e^{-x^n}dx$ converges $\iff n$ is even

Question

I'd like to show that $$\int_{-\infty}^{\infty}e^{-x^n}dx \text{ converges} \iff n\text{ is even.}$$

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My attempt:

Suppose $n$ is odd. Then $$\int_{-\infty}^{\infty}e^{-x^n}dx=\int_{0}^{\infty}e^{-x^n}dx+\underbrace{\int_{0}^{\infty}e^{x^n}dx}_{\text{diverges}}$$ The second integral diverges since $e^{x^n}\to\infty$ as $x\to\infty$.

If instead, $n$ is even, then $$\int_{-\infty}^{\infty}e^{-x^n}dx =2\int_{0}^{\infty}e^{-x^n}dx$$ I don't know how to show this is bounded.

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I would like to know if there is any inequality I can use to bound this, and show that the integral is finite.

Answer 1

Note that $x^n \le e^{x^n}$, so $\frac1{x^n}\ge e^{-x^n}$. With $f(x) = e^{-x^n}$, write:

$$\int_0^{\infty} f(x)dx = \int_0^1 f(x)dx + \int_1^{\infty}f(x)dx$$

The first integral on the RHS is obviously convergent. The second integral converges by comparison test.

Answer 2

HINT Clearly, $e^{x^4} > e^{x^2}$ so $$ \int_0^\infty \frac{dx}{e^{x^4}} < \int_0^\infty \frac{dx}{e^{x^2}} = J $$ and RHS is computable exactly since $$ J^2 = \left(\int_0^\infty \frac{dx}{e^{x^2}}\right) \left(\int_0^\infty \frac{dy}{e^{y^2}}\right) = \int_0^\infty \int_0^\infty \frac{dxdy}{e^{x^2+y^2}} = \frac{\pi}{2} \int_0^\infty \frac{rdr}{e^{r^2}} $$ which is now trivially integratable.

Answer 3

Hint:

$$0\le\int_0^\infty e^{-x^n}\mathrm dx\le\sum_{k=0}^\infty e^{-k^n}$$

Applying the root test we have that

$$\limsup_{k\to\infty}\sqrt[k]{e^{-k^n}}=\limsup_{k\to\infty}e^{-k^{(n-1)}}=0,\quad n>1$$