# Convergence of integral

Hello, it's me again ;)

Problem:
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Define

$$f_{n}(x)=n^{\alpha}, |x|\leq 1/n, f_{n}=0$$ elsewhere

Give all $$\alpha \in \Re$$ for which

$$\lim_{n \to \infty} \int_{\Re}f_{n}(x)dx=+\infty$$
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Can i change this last integral to:

$$\lim_{x \to 0} \int_{0}^{\infty} x^{-\alpha}dx=+\infty$$

But i think the integration limits aren't correct, and therefore $$alpha$$ is wrong too.

Any help appreciated,

kind regards,

W.

rachmaninoff
Theraven1982 said:
Hello, it's me again ;)

Problem:
-------
Define

$$f_{n}(x)=n^{\alpha}, |x|\leq 1/n, f_{n}=0$$ elsewhere

Give all $$\alpha \in \Re$$ for which

$$\lim_{n \to \infty} \int_{\Re}f_{n}(x)dx=+\infty$$
Let's see, your function f_n is a constant n^a on the interval [-1/n, 1/n], and zero elsewhere. So the integral is just

$$\int_{\Re}f_{n}(x)dx= n^a \cdot \frac{2}{n}= 2 n^{a-1}$$

as you can easily see from the graph of f_n.

You can take it from there...

Off course... now that I see it, it's all very simple. Guess sometimes my mind gets confused after too much maths ;).
Thank you,

W.