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I'm trying to get my head around calculating $$ \int\frac{1}{x\ln(x+1)}dx. $$ I can't seem to get anywhere. I tried parts and substitutions, but that $(x+1)$ is always in the way.

Any suggestions?

EDIT:
Thanks to all those who have pointed out that there is no simple expression for the integral. Does anyone know a closed-form approximation? I'm not interested in a definite integral, but rather in an approximation to $$ \int_1^x\frac{1}{t\ln(t+1)}dt. $$

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    $\begingroup$ This integral is actually not computable if you check Wolfram Mathematica. $\endgroup$ – NasuSama Mar 15 '14 at 16:35
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    $\begingroup$ Mathematica 8 returns this unevaluated. $\endgroup$ – Jason Zimba Mar 15 '14 at 16:36
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    $\begingroup$ Fairly often an antiderivative cannot be expressed in terms of the standard elementary functions. Some evidence that this might be the case in particular instance can be obtained by asking Wolfram Alpha, or Mathematica, or Maple to do it. $\endgroup$ – André Nicolas Mar 15 '14 at 16:36
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    $\begingroup$ Are you looking for a numerical approximation over a given interval? $\endgroup$ – Jason Zimba Mar 15 '14 at 16:36
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    $\begingroup$ I suppose $\int_1^x \frac{1}{t \ln(t+1)} \; dt$ is not a satisfactory answer here. $\endgroup$ – wckronholm Mar 15 '14 at 16:39
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Let $u=\ln(x+1)$ ,

Then $x=e^u-1$

$dx=e^u~du$

$\therefore\int\dfrac{1}{x\ln(x+1)}dx$

$=\int\dfrac{e^u}{u(e^u-1)}du$

$=\int\dfrac{1}{u(1-e^{-u})}du$

$=\int\sum\limits_{n=0}^\infty\dfrac{(-1)^nB_nu^{n-2}}{n!}du$ (with the formula in http://en.wikipedia.org/wiki/Bernoulli_number#Generating_function)

$=\int\left(\dfrac{1}{u^2}-\dfrac{1}{2u}+\sum\limits_{n=2}^\infty\dfrac{(-1)^nB_nu^{n-2}}{n!}\right)du$

$=-\dfrac{1}{u}-\dfrac{\ln u}{2}+\sum\limits_{n=2}^\infty\dfrac{(-1)^nB_nu^{n-1}}{n!(n-1)}+C$

$=-\dfrac{1}{\ln(x+1)}-\dfrac{\ln\ln(x+1)}{2}+\sum\limits_{n=2}^\infty\dfrac{(-1)^nB_n(\ln(x+1))^{n-1}}{n!(n-1)}+C$

$\therefore\int_1^x\dfrac{1}{t\ln(t+1)}dx$

$=\left[-\dfrac{1}{\ln(t+1)}-\dfrac{\ln\ln(t+1)}{2}+\sum\limits_{n=2}^\infty\dfrac{(-1)^nB_n(\ln(t+1))^{n-1}}{n!(n-1)}\right]_1^x$

$=\dfrac{1}{\ln2}-\dfrac{1}{\ln(x+1)}+\dfrac{\ln\ln2}{2}-\dfrac{\ln\ln(x+1)}{2}+\sum\limits_{n=2}^\infty\dfrac{(-1)^nB_n((\ln(x+1))^{n-1}-(\ln2)^{n-1})}{n!(n-1)}$

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Choose a number $N$ that is relatively large, like 10 or 20 or what have you, and then you might try $$\int_1^x{{dt\over t \ln\!{(t+1)}}} \approx \int_1^N{dt\over t\ln\!(t+1)} + \int_N^x{dt\over t\ln t}\qquad \qquad (x \geq N)$$ The first integral on the right-hand side can be evaluated numerically given the choice of $N$, and the second integral is $\ln\!{(\ln{x}/\ln{N})}$.

So for example, the choice $N = 20$ leads to the following approximation for $x \geq 20$: $$\int_1^x{{dt\over t \ln\!{(t+1)}}} \approx 0.9415 + \ln\!{(\ln{x}})\,.$$

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