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I am trying to solve the following improper integral $I$.

$$I = \int\limits_0^\infty\frac{1}{\left(x+1\right)\left(ax+1\right)\left(bx+1\right)} dx$$

However, I myself get the divergent answer. When I tried it through an online-calculator, it gives me something else. Can someone guide me through the following? I will really appreciate.

\begin{align*} I &=\lim_{x\to \infty}\frac{\left(a-1\right)b\ln\left(\left|bx+1\right|\right)+\left(a-ab\right)\ln\left(\left|ax+1\right|\right)+\left(b-a\right)\ln\left(\left|x+1\right|\right)}{\left(a-1\right)\left(b-1\right)\left(b-a\right)} \\ &\quad - \lim_{x\to 0}\dfrac{\left(a-1\right)b\ln\left(\left|bx+1\right|\right)+\left(a-ab\right)\ln\left(\left|ax+1\right|\right)+\left(b-a\right)\ln\left(\left|x+1\right|\right)}{\left(a-1\right)\left(b-1\right)\left(b-a\right)}. \end{align*}

While solving the part for $\lim_{x\to0}$, I get $0$ as the output, but if I solve $\lim_{x\to\infty}$, I get a divergent term. While in actual if I solve it through an online calculator, I get the following answer.

$$I = \dfrac{\left(a-1\right)b\ln\left(b\right)-a\ln\left(a\right)b+a\ln\left(a\right)}{\left(a-1\right)\left(b-1\right)\left(b-a\right)}$$

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    $\begingroup$ I edited the question a bit so that the title better reflects your question and the equations are more readable. Hope this is to your linking. $\endgroup$ Commented Feb 14, 2019 at 22:46

4 Answers 4

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} I & = \int_{0}^{\infty}{\dd x \over \pars{x + 1}\pars{ax + 1}\pars{bx + 1}} \\[5mm] & = \int_{0}^{\infty}\braces{2\int_{0}^{1}\dd u_{1}\int_{0}^{u_{1}}\dd u_{2}\,{1 \over \bracks{x + 1 + u_{1}\pars{a - 1}x + u_{2}\pars{b - a}x}^{\, 3}}}\dd x \\[5mm] & = 2\int_{0}^{1}\dd u_{1}\int_{0}^{u_{1}}\dd u_{2}\int_{0}^{\infty} {\dd x \over \braces{\bracks{\pars{a - 1}u_{1} + \pars{b - a}u_{2} + 1}x + 1}^{\, 3}} \\[5mm] & = \int_{0}^{1}\dd u_{1}\int_{0}^{u_{1}}\dd u_{2}\, {1 \over \pars{a - 1}u_{1} + \pars{b - a}u_{2} + 1} \\[5mm] & = \left.{1 \over b - a}\int_{0}^{1}\dd u_{1}\, \ln\pars{\bracks{a - 1}u_{1} + \bracks{b - a}u_{2} + 1} \,\right\vert_{\ u_{2}\ =\ 0}^{\ u_{2} =\ u_{1}} \\[5mm] & = {1 \over b - a}\int_{0}^{1}\dd u_{1}\braces{\vphantom{\Large A}% \ln\pars{\bracks{b - 1}u_{1} + 1} - \ln\pars{\bracks{a - 1}u_{1} + 1}} \\[5mm] & = {1 \over b - a}\braces{% \bracks{-1 + {b\ln\pars{b} \over b - 1}} - \bracks{-1 + {a\ln\pars{a} \over a - 1}}} \\[5mm] & = \bbx{{\pars{b - 1}a\ln\pars{a} - \pars{a - 1}b\ln\pars{b} \over \pars{a - 1}\pars{a - b}\pars{b - 1}}} \end{align}

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  • $\begingroup$ I know it is very late asking about it, but can you please explain about the first step in which you have to take $u_1$ and $u_2$ ? $\endgroup$
    – SJa
    Commented Mar 26, 2019 at 22:31
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    $\begingroup$ @Kashan No. It's not too late. It's still fine. I'm sorry I forget to mention that the involved identity is some of the Feynman Parametrizations. Thanks. $\endgroup$ Commented Mar 27, 2019 at 19:19
  • $\begingroup$ Thanks. Indeed it is very helpful. I really appreciate your help. $\endgroup$
    – SJa
    Commented Mar 28, 2019 at 6:58
  • $\begingroup$ Can I use your proof in my thesis, what will be the best way to give your reference. Can I cite your answer as the reference? I am afraid, my supervisor won't allow this. $\endgroup$
    – SJa
    Commented Apr 4, 2019 at 1:25
  • $\begingroup$ I am even happy to cite your actual name as well if you like (not sure if Felix Marin is your actual name) $\endgroup$
    – SJa
    Commented Apr 4, 2019 at 1:28
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Repeating the steps you probably did $$\frac{1}{(x+1)(ax+1)(bx+1)}=\frac{a^2}{(a-1) (a-b) (a x+1)}+\frac{b^2}{(b-1) (b-a) (b x+1)}+\frac{1}{(a-1) (b-1) (x+1)}$$ Now, assuming $a>0$, $b>0$ and $a\neq b$, consider $$I(p)=\int_0^p\frac{dx}{(x+1)(ax+1)(bx+1)} $$ $$I(p)=\frac{(a-b) \log (p+1)+a (b-1) \log (a p+1)-(a-1) b \log (b p+1)}{(a-1) (b-1) (a-b)}$$ Now, let $p=\frac 1 t$ and use Taylor series to get $$I\left(\frac 1t\right)=\frac{a (b-1) \log (a)-b(a-1) \log (b)}{(a-1) (b-1) (a-b)}-\frac{t^2}{2 a b}+O\left(t^3\right)$$ and, back to $p$ $$I(p)=\frac{a (b-1) \log (a)-b(a-1) \log (b)}{(a-1) (b-1) (a-b)}-\frac{1}{2 a bp^2}+O\left(\frac{1}{p^3}\right)$$

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  • $\begingroup$ So in the final step, $\large \frac{1}{2abp^3} \approx 0 $ and $O\Big(\frac{1}{p^3}\Big) \approx 0$ since $p \to \infty$ $\endgroup$
    – SJa
    Commented Feb 16, 2019 at 0:42
  • $\begingroup$ @Kashan. Perfectly correct ! $\endgroup$ Commented Feb 16, 2019 at 3:08
  • $\begingroup$ Thanks, by the way, I have seen $O(.)$ terms in many calculus-based solutions, but I never understand how they come from. I understand that they are big-O notation in algorithms, but how do they appear in calculus? $\endgroup$
    – SJa
    Commented Feb 16, 2019 at 17:03
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The estimation of the $\lim_{\infty}$ part is more subtle that you seem to have thought.

Remove the useless absolute values, take the divergent part wrt $x$ (ie some $\ln{x}$) apart from the rest and enjoy seeing its coefficient vanish. The rest will be a linear combination of logarithms of convergent functions of $x$ ($1+1/x$, $b+1/x$, and so on).

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As pointed out by other user, you must take cancellations into consideration when evaluating the limit $x\to\infty$. Hints for such cancellation can be glimpsed from the fact that (1) $\log(px+q) \sim \log x$ as $x\to\infty$ for $p > 0$ and that (2) the sum of `coefficients' is zero: $(a-1)b + (a-ab) + (b-a) = 0$.

So here is an actual computation: Using the fact that

$$(a-1)b \log x + (a-ab) \log x + (b-a) \log x = 0, $$

we find that

\begin{align*} &(a-1)b \log(bx+1) + (a-ab) \log(ax+1) + (b-a) \log (x+1) \\ &=(a-1)b [\log(bx+1) - \log x] + (a-ab) [\log(ax+1) - \log x] + (b-a) [\log (x+1) - \log x] \\ &= (a-1)b \log\left(b+\frac{1}{x}\right) + (a-ab) \log\left(a+\frac{1}{x}\right) + (b-a) \log \left(1+\frac{1}{x}\right) \\ &\xrightarrow[x\to\infty]{} (a-1)b \log b + (a-ab) \log a + \underbrace{(b-a) \log 1}_{=0} \end{align*}

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  • $\begingroup$ Sorry, but I cannot understand why sum is coefficients is Zero? Other than that I can understand what you have done. Thanks $\endgroup$
    – SJa
    Commented Feb 14, 2019 at 23:41
  • $\begingroup$ @Kashan, Is your question about the equality $(a-1)b + (a-ab) + (b-a) = 0$? $\endgroup$ Commented Feb 15, 2019 at 6:01
  • $\begingroup$ No. My query relates to the part of integration's solution when $x\to\infty$. In my understanding, it should be divergent, but I have tried a couple of online calculators and they give a convergent solution as mentioned above. $\endgroup$
    – SJa
    Commented Feb 16, 2019 at 0:38

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