I need to find uniform convergence of integral $$I(a) = \int_{0}^{+\infty}\frac{x^aarctgx}{4+x^n}dx$$ where a from $$N=(0;n-1), n >2$$ I tried to do Weierstrass M-test. But I get $$I(a)<\int_{0}^{+\infty}\frac{d(x^n+4)}{4+x^n}dx$$ And this integral do not converges.How to find uniform convergence ?
uniform convergence
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integration
uniform-convergence
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0I don't understand why you mention uniform convergence, because there is no parameter $n \to +\infty$. – 2017-04-16
1 Answers
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As $\displaystyle \forall x \in [0,+\infty): 0 \leq \arctan(x) < \frac{\pi}{2}$ we conclude $$\displaystyle I(a) < \int_\limits{0}^{\infty} \frac{x^a}{4+x^n}dx < \int_\limits{0}^{\infty} \frac{x^a}{x^n}dx$$ The integral on the right converges since $a < n-1$ $$ \int_\limits{0}^{\infty} \frac{x^a}{x^n}dx < \int_\limits{0}^{\infty} \frac{x^{n-1}}{x^n}dx = \int_\limits{0}^{\infty} \frac{1}{x}dx$$