Convergence of integrals by examining the sum

Question

I am just looking for someone to tell me if my reasoning is correct. Let's look at the integral: $$\int^{\infty}_{0} \frac{x^a}{x + 1}dx$$ where $a$ is a real number. Is it valid to say that the integral will converge when the sum converges: $$ \int^{\infty}_{0} \frac{x^a}{x + 1}dx \sim \sum^{\infty}_{n = 0} \frac{n^a}{n + 1}$$ So:

• for $a < 0$ the first term is undefined
• for $a=0$ the sum diverges
• for $a>1$ the sum diverges
• for $0<a<1$ the sum converges

Is it valid to conclude that the integral converges only for $0<a<1$? Does this reasoning work for other integrals of this type, is there something I should pay attention to?

Answer 1

No it is not a valid method in general, for the convergence we need to show that for some convergent $\sum a_n$

$$0\le \int^{\infty}_{0} \frac{x^a}{x + 1}dx \le\sum^{\infty}_{n = 0} a_n$$

As an alternative note that for $a\ge0$

$$\frac{x^a}{x + 1}\sim x^{a-1}=\frac1{x^{1-a}}$$

and for $a<0$ let $b=-a>0$

$$\frac{x^a}{x + 1}=\frac{1}{x^b(x + 1)}$$

and

• as $x\to 0^+ \implies \frac{1}{x^b(x + 1)} \sim \frac1{x^b}$

• as $x\to \infty \implies \frac{1}{x^b(x + 1)} \sim \frac1{x^{b+1}}$

then refer to limit comparison test to conclude that the given integral converges for $a\in(-1,0)$.

Answer 2

No, you cannot assert that the integral converges for $0 < a < 1$. Using comparison test you would notice that $$ \frac {x^a}{1+x} \sim \frac 1{x^{1-a}} \quad [n \to \infty], $$ and $\int_A^\infty (1/ x^p) \mathrm dx \;[A>0]$ converges iff $p > 1$. Hence the integral converges when and only when $\color{red}{a < 0}$ [WRONG HERE, see UPDATE 2].

Meanwhile, the series is also diverges by similar testing method.

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There is a test for convergence of series: integral test. However this requires the function be decreasing. So generally you cannot use the correspondent series to test the convergence for this kind of improper integrals [although I do not know the counterexample when $f$ is not decreasing. Any comments about this are welcome].

UPDATE

Thanks to @RobertIsrael, the integral test is applicable to your question since $f(x) = x^a /(x+1)$ is monotonic for sufficiently large $x$.

UPDATE 2

I was wrong in the comparison test above. When $a < 0$, the point $0$ would also be questionable. Combining the asymptotic behavior as $x \to 0^+$ and $x \to +\infty$, the conclusion should be: the integral converges iff $\boldsymbol {0 < a < 1}$.