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If you look at functions of the form $1/x^k$, $k>0$, it seems you can't have your cake and eat it too. If the integral of $1/x^k$ converges on $[1,\infty[$, then it diverges on $]0,1]$ and vice-versa, and if you try to balance things by picking $1/x$, your greediness is punished by having a function whose integral doesn't converge on both limits.

My question: can you construct a smooth function $f(x)$ that has a vertical asymptote at $x=0$ such that the integral

$$\int_{0}^{\infty} f(x) dx $$

converges?

What if I demand absolute convergence?

user46242
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2 Answers2

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Use for example $\frac{1}{\sqrt{x}}e^{-x}$.

Remark: I prefer to note that we can smoothly splice two smooth functions together using a bump function defined over an arbitrarily short interval.

André Nicolas
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  • How would you use a bump function to construct an example for this problem? – geometricK Jul 16 '16 at 14:00
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    @ougoah: There is a bump function $b(x)$ which is $1$ if $x\le 1$, $0$ for $x\ge 2$, and between $0$ and $1$ everywhere in the interval $(1,2)$. For $x\gt 0$, let $f(x)=\frac{1}{\sqrt{x}}b(x)$. – André Nicolas Jul 16 '16 at 15:05
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you can take $f(x) =e^{-x} .$