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I am required to find the values of $C$ for which the integral $$\int_0^{\infty}\frac{7x}{x^2+1}-\frac{7C}{3x+1}dx$$ converges.

I know by experimentation that it converges when $C=3$. I am, however, unable to show this in a rigorous way.

I get stuck when I need to evaluate $$\lim_{x\rightarrow\infty}(\frac{7}{2}\ln(x^2+1)-\frac{7C}{3}ln(3x+1))$$

Any help would be appreciated.

Taking the hint from DonAntonio, I wrote the following solution. Let me know if there is anything wrong with the way I expanded DonAntonio's solution.

$$\frac{7x}{x^2+1}-\frac{7C}{3x+1}=7\cdot\,\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}$$

When $C=3$ $$7\cdot\,\int_0^{\infty}\frac{(3-3)x^2+x-3}{(x^2+1)(3x+1)}=7\cdot\,\int_0^{\infty}\frac{x-3}{(x^2+1)(3x+1)}dx<\int_0^{\infty}\frac{x}{x^3}dx = \int_0^{\infty}\frac{1}{x^2} dx$$ Which is convergent.

When $C\neq3$ $$7\cdot\,\int_0^{\infty}\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}=7\cdot\,\int_0^{\infty}\frac{(3-C)x^2}{(x^2+1)(3x+1)}+7\cdot\,\int_0^{\infty}\frac{x-C}{(x^2+1)(3x+1)}dx$$

$$=|7\cdot\,\int_0^{\infty}\frac{(3-C)x^2}{(x^2+1)(3x+1)}dx+K| > \int_0^{\infty}\frac{x^2}{x^4} dx = \int_0^{\infty}\frac{1}{x^2} $$ For some constant $K$

Which is divergent.

niobe
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  • Find an equivalent of the integrand at $+\infty$ by reducing to same denominator...

    Or find a necessary condition for your primitive to not have an infinite limit. Hint : $\ln(x^2+1)\sim 2\ln x$...

     – Nicolas FRANCOIS Oct 24 '16 at 13:11
  • Collecting the logs together, the numerator behaves like $x^{7}$, so we want to know the values of $C$ for which $(3x+1)^{7C/3}$ behaves like $x^{7}.$ The binomial expansion behaves like its leading term, so the answer appears to be $C=3,$ but I can't even attempt anything more rigorous than this, at least right now. – Will R Oct 24 '16 at 13:16

2 Answers2

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Hints:

$$\frac{7x}{x^2+1}-\frac{7C}{3x+1}=7\cdot\,\frac{(3-C)x^2+x-C}{(x^2+1)(3x+1)}$$

Now, if $\;C\neq3\;$ then comparison with some multiple of $\;\frac1x\;$ will give you divergence...

DonAntonio
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You know an antiderivative, namely $$ \frac{7}{6}\bigl(3\ln(x^2+1)-2C\ln(3x+1)\bigr)= \frac{7}{6}\ln\frac{(x^2+1)^3}{(3x+1)^{2C}}= \frac{7}{6}\ln \frac{\bigl(1+\frac{1}{x^2\mathstrut}\bigr)^3x^6} {(3+\frac{1}{x})^{2C}x^{2C}} $$ It's quite easy now to see when the limit is finite.

On the other hand, doing an asymptotic comparison with a power of $1/x$ is easier.

egreg
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