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So I have an improper integral:

$$\int_{1}^{\infty} \frac{1}{x^a(x-1)^b} dx$$

I have to find for which $a,b\in \mathbb{R}$ it converges.

So my initial step is:

$$\int_{1}^{\infty} \frac{1}{x^a(x-1)^b} \leq ?\int_{1}^{\infty} \frac{1}{x^ax^b}dx$$

But since $x>1$ this does not seem to be correct, should i use this:

$$\int_{1}^{\infty} \frac{1}{x^a(x-1)^b} \leq \int_{1}^{\infty} \frac{1}{(x-1)^a(x-1)^b}dx$$

It gives me the same result: $a+b>1$; is this the correct way or am I missing something? Any help would be appreciated.

Bernard
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MathIsTheWayOfLife
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1 Answers1

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Hint: Divide

$$\int_{1}^{\infty}\dfrac{1}{x^{a}\left(x-1\right)^{b}}{\rm d}x=\int_{1}^{2}\dfrac{1}{x^{a}\left(x-1\right)^{b}}{\rm d}x+\int_{2}^{\infty}\dfrac{1}{x^{a}\left(x-1\right)^{b}}$$

and study the convergence of each integral separately.

eranreches
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  • why exactly $(1,2)$ and $(2,\infty)$? – MathIsTheWayOfLife Jun 17 '19 at 15:10
  • You can take whatever $L$ in the middle, i.e. $\left(1,L\right)$ and $\left(L,\infty\right)$. The idea is just to separate the integral in such a way it is improper because of one limit only. The LHS integral is problematic at $x=1$ and $x\rightarrow \infty$. In the RHS, the first integral has a problem at $x=1$ and the left one has problem at $x\rightarrow\infty$. Now you can treat them separately. – eranreches Jun 17 '19 at 15:12
  • Oh okay, thank you very much – MathIsTheWayOfLife Jun 17 '19 at 15:15
  • To proceed, check https://services.math.duke.edu/~cbray/Stanford/2003-2004/Math%2042/limitcomp.pdf. – eranreches Jun 17 '19 at 15:17