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I have this exercise, and I get the right result. But while I think the first part is ok, the second part is from a formal stand point pretty hairy. So here the first part (I left some step out, but they should be for the most of you obvious):

$$\int_1^\infty \frac{1}{x+x^3} \,dx = \int_1^\infty \frac{1}{x(1+x^2)} \,dx= \int_1^\infty \frac{1}{x}-\frac{x}{(1+x^2)} \,dx= \left[\ln(x)\right]_1^\infty - \left[\frac{1}{2} ln(1+u)\right]_1^\infty$$

In the following second part, at least when I start to do arithmetic manipulation with $\infty$, it's not formal anymore.

$$ \left[\ln(x)\right]_1^\infty - \left[\frac{1}{2} ln(1-u)\right]_1^\infty = \ln(\infty)-ln(1)-\frac{1}{2} \ln(1-\infty)+\frac{1}{2}\ln(2) = \infty -\frac{1}{2} \infty +\frac{1}{2}\ln(2) = \frac{\ln 2}{2}$$

So it would be great if someone could show me a formal way to solve this integral.

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$\int_1^\infty \frac{1}{x}-\frac{x}{(1+x^2)} \,dx= \left[\ln(x)\right]_1^\infty - \left[\frac{1}{2} ln(1+u)\right]_1^\infty$ that step is wrong because that's only true when both the integrals on the RHS are finite, which isn't the case. Note that $\int_1^\infty \frac{1}{x} dx$ diverges. –  Git Gud Jan 14 '13 at 20:34
    
Thanks. Could fix that with $lim_{a->\infty} \int_1^a \, dx$, right? –  leo Jan 14 '13 at 20:37
    
No, you'd still be dealing with $+\infty$ and $-\infty$. Unless you mean keeping the whole thing on one integral as people are doing below. That's allowed. You can't, however, compute the integral for each function seperatly. –  Git Gud Jan 14 '13 at 20:39
    
That's what I meant, I should have written it a bit more longish. –  leo Jan 14 '13 at 21:01

7 Answers 7

up vote 7 down vote accepted

An improper integral is simply a limit of proper integrals:

$$\int_1^\infty\frac1{x+x^3}dx=\lim_{a\to\infty}\int_1^a\left(\frac1x-\frac{x}{1+x^2}\right)dx\;.$$

Keep the limits until you’re done finding the antiderivatives, and keep the pieces together:

$$\begin{align*} \int_1^\infty\frac1{x+x^3}dx&=\lim_{a\to\infty}\int_1^a\left(\frac1x-\frac{x}{1+x^2}\right)dx\\\\ &=\lim_{a\to\infty}\left[\ln x-\frac12\ln\left(1+x^2\right)\right]_1^a\\\\ &=\lim_{a\to\infty}\left[\ln x-\ln\left(1+x^2\right)^{1/2}\right]_1^a\\\\ &=\lim_{a\to\infty}\left[\ln\frac{x}{\left(1+x^2\right)^{1/2}}\right]_1^a\\\\ &=\lim_{a\to\infty}\ln\frac{a}{\left(1+a^2\right)^{1/2}}-\ln 2^{-1/2}\\\\ &=\frac12\ln 2\;, \end{align*}$$

since $\dfrac{a}{\left(1+a^2\right)^{1/2}}\to 1$ as $a\to\infty$.

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for me $-\ln 2^{-1/2} \neq \frac{1}{2} \ln2$. do i do a calculation mistake? –  leo Jan 14 '13 at 21:16
    
@leo: $\ln a^b=b\ln a$, so $-\ln 2^{-1/2}=-\left(-\frac12\right)\ln 2=\frac12\ln 2$. –  Brian M. Scott Jan 14 '13 at 21:20
    
yes, thanks! should have seen that one! –  leo Jan 14 '13 at 21:22
    
@leo: You’re welcome! –  Brian M. Scott Jan 14 '13 at 21:27

Substitute $x = \tan{\theta}$, $dx = \sec^2{\theta} d \theta$:

$$\int_1^{\infty} \frac{dx}{x (1+x^2)} = \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} d \theta \: \cot{\theta} = \left [ \log{\sin{\theta}} \right ]_{\frac{\pi}{4}}^{\frac{\pi}{2}} = -\log{\frac{1}{\sqrt{2}}} $$

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This seams like a very nice solution to me. Unfortunately I do not (yet) have the Math skills to follow it/understand it fully. –  leo Jan 14 '13 at 21:21
    
I'll fill in a little here. When $x = \tan{\theta}$, $1+x^2 = 1 + \tan^2{\theta} = \sec^2{\theta}$. Note that there is a factor of $\sec^2{\theta}$ in both numerator and denominator, so they cancel. All you are left with is the factor of $x = \tan{\theta}$ in the denominator, which gives you the $\cot{\theta}$ in the second step. The integration limits come from the fact that $\theta = \arctan{x}$, and $\arctan{\infty} = \pi/2$ and $\arctan{1} = \pi/4$. Let me know what else I can fill in. –  Ron Gordon Jan 14 '13 at 21:35

Set $x = \dfrac1u$. Hence, $dx = - \dfrac{du}{u^2}$. This gives us $$I = \int_1^{\infty} \dfrac{dx}{x+x^3} = \int_0^1 \dfrac{du}{u^2 \left(\dfrac1u + \dfrac1{u^3} \right)} = \int_0^1 \dfrac{u du}{1+u^2} = \left. \dfrac12 \log(1+u^2) \right \vert_{u=0}^{u=1} = \dfrac{\log 2}2$$

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Marvis, I am impressed with your consistently clever approach to these problems (clever = "ways I didn't come up with"). I would not have picked this way to solve the problem, but I like how the basic result falls out using simple functions. –  Ron Gordon Jan 14 '13 at 21:37
    
@rlgordonma Thanks for your appreciation. –  user17762 Jan 14 '13 at 22:46

Use instead $$\int\frac1{x(1+x^2)}\mathrm dx=\log\varphi(x),\qquad\varphi(x)=\frac{x}{\sqrt{1+x^2}}, $$ and note that $\varphi(1)=\frac1{\sqrt2}$ and $\varphi(+\infty)=1$.

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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ \begin{align} &\overbrace{\color{#66f}{\large\int_{1}^{\infty}{\dd x \over x + x^{3}}}} ^{\ds{x \equiv \expo{t}}}\ =\ \int_{0}^{\infty}{\expo{-2t} \over 1 + \expo{-2t}}\,\dd t =\left.{\ln\pars{1 + \expo{-2t}} \over -2}\,\right\vert_{\, 0}^{\,\infty} =\color{#66f}{\large\half\,\ln\pars{2}} \end{align}

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We'll pick up where you wrote $$\int_1^\infty\frac{1}{x}-\frac{x}{(1+x^2)}dx.$$ Note that we have $$\lim_{N\to\infty}\left[\ln(x)-\frac{1}{2}\ln(1+x^2)\right]_{1}^N=\lim_{N\to\infty}\left[\ln(N)-\frac{1}{2}\ln(1+N^2)-\left(\ln(1)-\frac{1}{2}\ln(2)\right)\right].$$ Simplifying the natural logs, we have $$\lim_{N\to\infty}\left[\ln\left(\frac{N}{\sqrt{1+N^2}}\right)+\frac{\ln(2)}{2}\right].$$ It is easy to see that the argument of $\ln$ tends to $1$, thus the first term goes to $0$. This leaves us with the desired answer.

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Letting $x=\sqrt u$, our integral becomes $$\frac{1}{2}\int_1^{\infty}\frac{1}{u(u+1)} \mathrm{du}=\left[\frac{1}{2}\ln \left(\frac{u}{u+1}\right)\right]_1^{\infty}=\frac{\ln 2}{2}$$

Chris.

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