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For, example, question - is $\int_0^{\infty} \frac{x}{x+1}$ convergence

Only one place, where we interesting in this integral is ($\infty$).

We can take function $g(x)$, that $\lim\limits_{x \rightarrow \infty} \frac{\frac{x}{x + 1}}{g(x)} = n \in \mathbb{Z} \ne 0$.

And if we know about convergence in $\infty \int g(x)$, the convergence of our integral will be same.

But by the rule we can take only $g(x)$ that $\forall x$ will be $>0$.

Why i can't say that $\int_{0}^{\infty}\frac{\cos(x)x}{x + 1}$ not convergence, because is no limit $x \rightarrow \infty \ F(cos(x))$

Eugene Korotkov
  • 409

2 Answers2

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Note that

$$\int_0^c \frac{x}{x+1} \, dx = \int_0^c \frac{x+1}{x+1} \, dx - \int_0^c \frac{1}{x+1} \, dx = c - \ln (c+1) = \ln \left(\frac{e^c}{c+1} \right),$$

and

$$\lim_{c \to \infty}\int_0^c \frac{x}{x+1} \, dx = +\infty$$

Since the integrand here is nonnegative we also can show divergence using the limit comparison test with the "equivalent function" $g(x) = 1.$

In the second case, the limit comparison test is not useful because the integrand assumes both positive and negative values infinitely often. The integral $\int_0^\infty \frac{x \cos x }{x+1} \, dx $ fails to converge because

$$\int_0^c \frac{x \cos x}{x+1} \, dx = \int_0^c \cos x\, dx - \int_0^c \frac{\cos x}{x+1} \, dx ,$$

and the first integral on the RHS diverges while the second converges by the Dirichlet test.

Keep in mind that there are cases where $\int_0^\infty f(x) \, dx $ converges yet $\lim_{x \to \infty} f(x)$ does not exist. Failure of that limit to exist is not sufficient to prove that the improper integral diverges.

RRL
  • 90,707
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For $x\ge1$, we have that $\frac{x}{x+1}\ge\frac12$. This implies that $$ \int_1^L\frac{x}{x+1}\,\mathrm{d}x\ge\frac{L-1}{2} $$ Thus, $\int_0^\infty\frac{x}{x+1}\,\mathrm{d}x$ diverges.

Furthermore, since $\cos(x)$ has one sign over each interval $\left[\left(k-\frac12\right)\pi,\left(k+\frac12\right)\pi\right]$, $$ \begin{align} \left|\,\int_{\left(k-\frac12\right)\pi}^{\left(k+\frac12\right)\pi}\frac{x\cos(x)}{x+1}\,\mathrm{d}x\,\right| &\ge\frac12\left|\,\int_{\left(k-\frac12\right)\pi}^{\left(k+\frac12\right)\pi}\cos(x)\,\mathrm{d}x\,\right|\\[9pt] &=1 \end{align} $$ Therefore, $\int_0^\infty\frac{x\cos(x)}{x+1}\,\mathrm{d}x$ diverges.

robjohn
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