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finding $\displaystyle \int^{\infty}_{0}\frac{\ln x}{x^2+6x+9}dx$

Attempt: let $\displaystyle I(a) = \int^{\infty}_{0}\frac{\ln (ax)}{(x+3)^2}dx, a>0$

$\displaystyle I'(a) = \int^{\infty}_{0}\frac{x}{ax(x+3)^2}dx = \frac{1}{a}\int^{\infty}_{0}\frac{1}{(x+3)^2}dx = -\frac{1}{a}\bigg(\frac{1}{x+3}\bigg)\bigg|_{0}^{\infty} = \frac{1}{3a}$

so $\displaystyle I(a) = \frac{\ln(a)}{3}+C$

could some help me how to solve from there, thanks in advanced

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  • $\begingroup$ Why don't you integrate by parts? $\endgroup$ Feb 25, 2017 at 8:57
  • $\begingroup$ The problem with your approach is that when you substitute $a=1$ as necessary, you end up with the fact that $I(a) = c$, a constant, which we already knew. Very nice attempt, though. $\endgroup$ Feb 25, 2017 at 9:04
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    $\begingroup$ This method works if you know the integral for some value of $a$. It doesn't appear that we know this integral for any value of $a$. $\endgroup$
    – robjohn
    Feb 25, 2017 at 9:18
  • $\begingroup$ integrate $ \int^{\infty}_{0}\frac{\ln^2 x}{x^2+6x+9}dx$ around a keyhole in the complex plane $\endgroup$
    – tired
    Feb 25, 2017 at 10:28

3 Answers 3

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Hint. Try integration by parts. Take

$$u=\ln x \qquad \text{and} \qquad dv=\frac{dx}{(x+3)^2}.$$

Then, the integral is equal to

$$\int_{\frac{1}{M}}^M \frac{\ln x}{(x+3)^2}dx=\left[-\frac{\ln x}{x+3}\right]_{\frac{1}{M}}^M+\int_{\frac{1}{M}}^M \frac{dx}{(x+3)x}$$

You can solve the last integral applying the theory of rational integrals and take the limit $M \to \infty$.

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Hint. Alternatively integrating by parts, with $M>0$ and $\varepsilon>0$, one gets $$ \int^M_{\varepsilon}\frac{\ln x}{(x+3)^2}dx=\left[\left(\frac13-\frac{1}{(x+3)}\right)\cdot \ln x\right]^M_{\varepsilon}-\int^M_{\varepsilon}\left(\frac13-\frac{1}{(x+3)}\right)\cdot \frac1x\:dx, $$ the latter integral is easier, then one may let $M \to \infty$, $\varepsilon \to 0^+$.

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  • $\begingroup$ I wrote out a full answer using $M=\infty$, but a hint is probably sufficient. (+1) One thing that should be mentioned is that $\epsilon\to0^+$ $\endgroup$
    – robjohn
    Feb 25, 2017 at 9:13
  • $\begingroup$ @robjohn Correct, even if we know that $\varepsilon>0$, it is preferable to mention it. Thank you. $\endgroup$ Feb 25, 2017 at 9:15
  • $\begingroup$ Sorry, I just noticed that you specified that $\epsilon\gt0$. $\endgroup$
    – robjohn
    Feb 25, 2017 at 9:16
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If $\displaystyle I=\int\limits_0^\infty \dfrac{\ln x}{(x+a)^2}\; dx$ for some $a\gt 0$,

Then the substitution $xy=a^2$ gives

$\displaystyle I=\int\limits_0^\infty \dfrac{2\ln(a)-\ln(y)}{(y+a)^2}\; dy\\2I=\displaystyle 2\ln a\int\limits_0^\infty \dfrac{dy}{(y+a)^2} = \dfrac{2\ln a}{a}$

Therefore, $\displaystyle \int\limits_0^\infty \dfrac{\ln x}{(x+a)^2}\; dx = \dfrac{\ln a}{a}$

For this one, the answer is $\boxed{\dfrac{\ln 3}{3}}$

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