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Consider the following integral: $$\int_{0}^{\infty}\frac{dx}{x^a+x^b}$$

What are the necessary (and sufficing) conditions on $a$ and $b$ for the integral to converge?

It's easy to see that if $a=b$, then the integral is necessarily divergent. But as soon as I start checking for $a\neq b$, I get lost in all the different cases.

virtualcode
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    near $x=0$, $x^a+x^b\sim x^{\min{a,b}}$ and as $x\to \infty$, $x^a+x^b\sim x^{\max{a,b}}$ – Sine of the Time Feb 25 '24 at 18:30
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    WLOG we may assume $a\le b.$ Then for $x\le 1$ we have $${1\over 2x^a}\le {1\over x^a+x^b}\le {1\over x^a}$$ while for $x>1$ there holds $${1\over 2x^b}\le {1\over x^a+x^b}\le {1\over x^b}$$ So $a<1$ and $b>1.$ – Ryszard Szwarc Feb 25 '24 at 18:42
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    Not an answer, but if you know that the common integral $\int_0^\infty \frac{x^{\mu-1}}{1+x^\nu}dx$ converges absolutely for $0<\mu<\nu$ then noticing that $\frac{1}{x^a + x^b}=\frac{x^{(1-a)-1}}{1+x^{b-a}}=\frac{x^{(1-b)-1}}{1+x^{a-b}}$ gives you convergence: if $a<b$ when $a<1<b$, and if $a>b$ when $a>1>b$. – Robert Lee Feb 25 '24 at 22:46
  • I checked with various suitable values for $a$ & $b$ : weirdly wolfram is not supporting the various conclusions in all the comments here & the answer too ! – Prem Feb 26 '24 at 13:00
  • @Prem what values of $a$ and $b$ did you consider? – Sine of the Time Feb 26 '24 at 18:56
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    Oh , Issue was in autocorect , autosuggestion & typo !! While these 2 did not work : https://www.wolframalpha.com/input?i=INTEG+%281%2F%28X%5E0.2%2BX%5E3%29%29+from+x%3D0+to+infinity https://www.wolframalpha.com/input?i=INTEG+%281%2F%28X%5E2%2BX%5E-3%29%29+from+x%3D0+to+infinity I now see what was wrong & these 2 work : https://www.wolframalpha.com/input?i=INTEG+%281%2F%28X%5E0.2%2BX%5E3%29%29+from+X%3D0+to+infinity https://www.wolframalpha.com/input?i=INTEG+%281%2F%28X%5E2%2BX%5E-3%29%29+from+X%3D0+to+infinity Hence it is a Non-Issue ! – Prem Feb 27 '24 at 06:51

1 Answers1

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WLOG suppose that $b\geqslant a$, then $$ \int_{0}^{\infty }\frac{dx}{x^a+x^b}=\int_{0}^{\infty }\frac{x^{-a}}{1+x^{b-a}}\,d x $$

Now observe that if $a\geqslant 1$ the integral diverges, as $\frac{x^{-a}}{1+x^{b-a}}\geqslant \frac{x^{-a}}2$ for $x\in(0,1]$ and $\int_{0}^1 x^{-a}\,d x=+\infty $. By the other hand, if $a< 1$ then the integral in $[0,1]$ is well-defined, and converges on $[1,\infty )$ if and only if $b>1$. This happens because for $x\geqslant 1$ you can write

$$ \frac{x^{-a}}{1+x^{b-a}}=\frac{x^{b-a}}{1+x^{b-a}}\cdot x^{-b} $$

and $0\leqslant \frac{x^{b-a}}{1+x^{b-a}}\leqslant 1$ for all $x\geqslant 0$. Therefore we conclude that the integral converges if and only if $\max\{ a,b \}>1$ and $\min\{ a,b \}<1$.∎

Masacroso
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  • I checked with some suitable values for $a$ & $b$ : wolfram is not supporting your conclusions – Prem Feb 26 '24 at 12:54
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    @Prem I checked my answer and it had a silly mistake that was fixed now. Anyway wolfram fails very much on analysis, so its a tool to use with care to check some results – Masacroso Feb 26 '24 at 13:08
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    Oh , I know that , which was why I did not claim your answer was wrong or that wolfram was right. Earlier , I was giving wrong inputs to wolfram , which is working fine & confirms your answer !! – Prem Feb 27 '24 at 07:00