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Say you have the integral $\displaystyle\int_1^\infty{\frac{1}{x^{1+\frac{1}{x}}}}\;\mathrm{d}x$

This integral cannot be completed. Not that it goes to infinity, but it physically just cannot be completed. How can you realize this if you encounter it? How can you prove it?

Kalcifer
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  • I believe by "physically just cannot be completed" you mean cannot be expressed in terms of elementary/standard functions (since you can approximate definite integrals numerically). By the way your integral does not converge. – player3236 Nov 20 '20 at 06:12
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    I don't know what you mean by saying that you cannot complete it. The integrand is a positive continuous function and the integral makes sense and is has the value $+\infty$. – Kavi Rama Murthy Nov 20 '20 at 06:12
  • @player3236 You are correct, I should have been more specific with what I said. I do mean elementary/standard functions. And correct it does not converge, but how do I know? – Kalcifer Nov 20 '20 at 06:14
  • @KaviRamaMurthy How do you surmize this if you cannot complete the integral to take its limit as x approaches infinity? The indefinite integral does not exist. – Kalcifer Nov 20 '20 at 06:15
  • You don't have to know the indefinite integral to show that the value is $\infty$. You can prove this by comaparison with the integral of $\frac 1 x$. – Kavi Rama Murthy Nov 20 '20 at 06:16
  • @KaviRamaMurthy Same idea as the comparison test with a series? – Kalcifer Nov 20 '20 at 06:17
  • @KaviRamaMurthy just gave me the answer. My problem has been solved thank you. Is the comparison test the only way that you can solve this type of question? – Kalcifer Nov 20 '20 at 06:18
  • Yes. To be sepcific you can show that $\frac 1{x^{1+\frac 1 x} } >\frac 1{2x}$ for $x$ sufficiently large. – Kavi Rama Murthy Nov 20 '20 at 06:19
  • You cannot compare it to $\frac1x$ directly. However for your integral, check out https://math.stackexchange.com/questions/737278/how-to-see-this-improper-integral-diverges and https://math.stackexchange.com/questions/1792839 – player3236 Nov 20 '20 at 06:19
  • @player3236 Why can you not compare it to $\frac{1}{x}$ directly? – Kalcifer Nov 20 '20 at 06:23
  • See this post and the links therein https://math.stackexchange.com/questions/2285346/check-if-antiderivative-is-elementary When applying Liouville's Theorem, it might be easier to convert first the integral via $x = \exp u$ to $\int \exp\left[-u \exp(-u)\right] ,du$. – Travis Willse Oct 16 '23 at 05:14

2 Answers2

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@KaviRamaMurthy and @player2326 have answered this question. The comparison test can be employed to solve this question.

Additional references:

How to see this improper integral diverges?

Checking whther the integral $\int_1^∞ \frac{1}{x^{\frac{1}{x}+1}} dx$ convergent

Kalcifer
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$$I=\int_{1}^{\infty}\frac{dx}{x^p}=\left .\frac{x^{1-p}}{1-p}\right|_{1}^{\infty}=\frac{1}{p-1}, ~if~ p>1,$$ because $0^{1-p}=0$ if only $p>1$, otherwise it is infinite. Hence the integral converges when $p>1$.

Z Ahmed
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