I don't get how we're supposed to use analysis to calculate things like:

a) $$ \int_0^1 \log x \mathrm dx $$ b) $$\int_2^\infty \frac{\log x}{x} \mathrm dx $$ c) $$\int_0^\infty \frac{1}{1+x^2} \mathrm dx$$

I tried integration by parts, but not helpful since we have it unbounded. Please explain how to do this using improper integrals in analysis.

  • 1
    $\begingroup$ For a), you could try the substitution $x=\exp(-u)$... b) looks divergent, and c) succumbs to the substitution $x=\tan(u)$. $\endgroup$ – J. M. isn't a mathematician Apr 28 '11 at 9:46
  • 1
    $\begingroup$ For a) Integrate by parts $\int_{0}^{1}\log x\;\mathrm{d}x=\int_{0}^{1}1% \cdot \log x\;\mathrm{d}x$. Apply l'Hôpital to $\lim_{x\rightarrow 0^{+}}x\log x$. $\endgroup$ – Américo Tavares Apr 28 '11 at 12:03
  • $\begingroup$ b) is divergent for sure, either by comparison to integrating $1/x$, or by by contemplating the derivative of $(\log x)^2$. $\endgroup$ – Gerry Myerson Apr 28 '11 at 13:08
  • $\begingroup$ One question per post please! $\endgroup$ – jimjim Jan 12 '15 at 12:15

Remember that $ \int_a^{\infty} f(x) \ dx = \lim_{b \to \infty} \int_a^b f(x) \ dx $.

For questions b) and c), you want to determine whether or not these limits exist (and if so, what they are). You can do this by integrating between $ a $ and $ b $ first, and then taking the limit as $ b \to \infty $ (you can integrate both using standard techniques).

For a), the substitution J.M. suggested will help (the substitution $ x = \exp{(u)} $ works just as well). This will convert the integral into an improper one, and you can then use the same method as for b) and c).

EDIT: In response to your request for an example, say we wanted to know whether or not $ \int_1^{\infty} \frac{1}{x} \ dx $ exists. Then we can first calculate $ \int_1^b \frac{1}{x} \ dx = \log{b} $, and note that $ \int_1^{\infty} \frac{1}{x} \ dx = \lim_{b \to \infty} \int_1^b \frac{1}{x} \ dx = \lim_{b \to \infty} \log{b} $. Does this limit exist?

  • $\begingroup$ You can also do a) using integration-by-parts, since $d/dx(xlog(x)) = log(x) + 1$ and then you have to look at whether $\lim_{x \to 0} (xlog(x))$ exists $\endgroup$ – Chris Card Apr 28 '11 at 11:10

Edited to write the improper limits as limits of proper integrals, in response to Arturo's comment.

a) The integrand $\log x$ has a singular point at $x=0$. The improper integral of the second king $I=\int_{0}^{1}\log x\;\mathrm{d}x$ is, by definition, the limit

$$\lim_{\varepsilon \rightarrow 0^{+}}\int_{\varepsilon }^{1}\log x\;\mathrm{% d}x.$$

The integral $\int \log x\;\mathrm{d}x$ is usually integrated by parts

$$\begin{eqnarray*} I &=&\lim_{\varepsilon \rightarrow 0^{+}}\int_{\varepsilon }^{1}\log x\;% \mathrm{d}x=\lim_{\varepsilon \rightarrow 0^{+}}\int_{\varepsilon }^{1}1\cdot \log x\;\mathrm{d}x \\ &=&\lim_{\varepsilon \rightarrow 0^{+}}\left[ x\log x\right] _{\varepsilon }^{1}-\lim_{\varepsilon \rightarrow 0^{+}}\int_{\varepsilon }^{1}x\cdot \frac{1}{x}\;\mathrm{d}x \\ &=&1\cdot \log 1-\lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \log \varepsilon -1 =-\lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \log \varepsilon -1=-1, \end{eqnarray*}$$

where $\lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \log \varepsilon $ was evaluated by l'Hôpital's rule:

$$\lim_{\varepsilon \rightarrow 0^{+}}\varepsilon \log \varepsilon =\lim_{\varepsilon \rightarrow 0^{+}}\frac{\log \varepsilon }{\frac{1}{% \varepsilon }}=\lim_{\varepsilon \rightarrow 0^{+}}\frac{\frac{1}{% \varepsilon }}{-\frac{1}{\varepsilon ^{2}}}=\lim_{\varepsilon \rightarrow 0^{+}}-\varepsilon =0.$$


b) The integral $\int_{a}^{+\infty }\frac{1}{x^{p}}\;\mathrm{d}x$ is divergent for $a>0,p\leq 1$, as can be seen by evaluating it. We apply the limit test to $f(x)=\frac{\log x}{x% }$ and $g(x)=\frac{1}{x}$:

$$\frac{f(x)}{g(x)}=\frac{\log x}{x}\cdot x=\log x\rightarrow \infty,\qquad\text{ as } x\rightarrow \infty .$$

Both $f(x)$ and $g(x)$ are nonnegative functions in $[2,+\infty \lbrack $. Since $\int_{2}^{\infty }g(x)\;\mathrm{d}x=\int_{2}^{\infty }\frac{1}{x}\;\mathrm{d}x$ is divergent, so is $\int_{2}^{\infty }f(x)\;\mathrm{d}x=\int_{2}^{\infty }\frac{\log x}{x}\;% \mathrm{d}x.$

c) The improper integral $I=\int_{0}^{\infty }\frac{1}{1+x^{2}}\;\mathrm{d}x$ is of the first kind, because the integrand has no singularities. By definition of an integral of such a kind, it is the limit

$$\lim_{b\rightarrow +\infty }\int_{0}^{b}\frac{1}{1+x^{2}}\;\mathrm{d}x.$$

Since $\int \frac{1}{1+x^{2}}\;\mathrm{d}x=\arctan x$, we have: $$\begin{eqnarray*} I &=&\lim_{b\rightarrow +\infty }\int_{0}^{b}\frac{1}{1+x^{2}}\;\mathrm{d}x =\lim_{b\rightarrow +\infty }\left[ \arctan x\right] _{0}^{b} \\ &=&\lim_{b\rightarrow +\infty }\arctan b-\arctan 0 =\frac{\pi }{2}-0=\frac{\pi }{2}. \end{eqnarray*}$$


Here's an alternative derivation for a), where improperness plays no role.

$$ \int_0^1 { - \log x\,{\rm d}x} = \int_{x = 0}^1 {\int_{u = x}^1 {\frac{1}{u}\,{\rm d}u} \,{\rm d}x} = \int_{u = 0}^1 {\int_{x = 0}^u {\frac{1}{u}\,{\rm d}x} \,{\rm d}u} = \int_{u = 0}^1 {\frac{1}{u}u\,{\rm d}u} = 1. $$


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