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Let $a_n = \int\limits_{0}^{n} e^{-x^4} dx$. Does $\{ a_n \}_{n \rightarrow \infty}$ converge?

$\{ a_n \} =\{ \int\limits_{0}^{1} e^{-x^4} dx, \int\limits_{0}^{2} e^{-x^4} dx, ..., \int\limits_{0}^{\infty} e^{-x^4} dx \}$

So we need need only to check that the definite integral

$\int\limits_{0}^{\infty} e^{-x^4} dx$ converges

By using Wolfram Alpha,

$\int\limits_{0}^{\infty} e^{-x^4} dx \} = \Gamma \left( \frac54 \right) \approx 0.906402$

where $\Gamma$ is the Gamma function

Therefore $\{ a_n \}$ converges, to $\Gamma \left( \frac54 \right)$.

But what if I can't use Wolfram Alpha? How can I solve for this integration by hand? Sorry if this makes me look stupid, I don't recall learning any techniques from Calculus that can help me solve this integration.

Thanks in advance!

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    $\begingroup$ Exponential are 'always' quite strong. $\endgroup$ – Felix Marin Oct 31 '13 at 7:47
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    $\begingroup$ Generally speaking, $$n!=G\left(\frac1n\right)\qquad where \qquad G(n)=\int_0^\infty e^{-x^n}dx\qquad\forall\ n\geqslant0$$ $\endgroup$ – Lucian Oct 31 '13 at 7:47
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We do not need an explicit expression to show that an improper integral converges.

The sequence $(a_n)$ is obviously increasing. It is bounded above by $\int_0^1 e^{-x^4}\,dx+\int_1^\infty e^{-x}\,dx$. To be explicit, the first integral is less than $1$, and the second is $e^{-1}$, so the sequence $(a_n)$ is bounded above by $1+e^{-1}$.

Any increasing sequence which is bounded above converges.

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  • $\begingroup$ You are welcome. $\endgroup$ – André Nicolas Nov 19 '13 at 3:07
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Recall the Taylor series for $e^x$ at $x = 0$ is given by:

\begin{equation} e^{x} = \sum_{n = 0}^{\infty} \frac{x^n}{n!} \end{equation}

Thus,

\begin{equation} e^{-x^4} = \sum_{n = 0}^{\infty} \frac{(-1)^n x^{4n}}{n!} \end{equation}

Thus,

\begin{align} \int_{0}^{t} e^{-x^4} \:dx &= \sum_{n = 0}^{\infty} \frac{(-1)^n }{n!}\int_{0}^{t}x^{4n}\:dx = \sum_{n = 0}^{\infty} \frac{(-1)^n }{n!} \left[\frac{x^{4n + 1}}{4n + 1} \right]_{0}^{t} \\ &= \sum_{n = 0}^{\infty} \frac{(-1)^n }{n!} \frac{t^{4n + 1}}{4n + 1} = \sum_{n = 0}^{\infty} b_n(t) \end{align}

We now apply the Ratio Test:

\begin{align} R &= \lim_{n \rightarrow \infty} \left|\frac{b_{n + 1}}{b_{n}}\right| = \lim_{n \rightarrow \infty} \left|\frac{\frac{(-1)^{n+1} }{\left(n + 1\right)!} \frac{t^{4\left(n + 1\right) + 1}}{4\left(n + 1\right) + 1}}{\frac{(-1)^n }{n!} \frac{t^{4n + 1}}{4n + 1}}\right| \\ &= \lim_{n \rightarrow \infty} \left| \frac{1}{n + 1} \cdot \frac{4n + 1}{4n + 5}\cdot t^4\right| \end{align}

Which for any finite $t$ becomes $R = 0$ and thus, by the Ratio test the series (and by extension) integral converge for all finite real $t$.

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