The indefinite integral $\int\frac{dx}{x\ln(x+1)}$

Question

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 $$ \int_1^x\frac{1}{t\ln(t+1)}dt. $$

Answer 1

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)

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

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

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

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

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

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

Answer 2

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 $x \geq 20$: $$\int_1^x{{dt\over t \ln\!{(t+1)}}} \approx 0.9415 + \ln\!{(\ln{x}})\,.$$

Answer 3

If you are interested in the asymptotics of the function $$ \int_1^x \frac{1}{t\log(1+t)} dt $$ for large $x$, one good way to approximate it is to observe that $\frac1t \approx \frac1{1+t}$. In fact $$ \frac1t = \frac1{1+t}+\frac1{t(1+t)} $$ so \begin{eqnarray} \int_1^x \frac{1}{t\log(1+t)} dt &=& \int_1^x \frac{1}{(1+t)\log(1+t)} dt + \int_1^x \frac{1}{t(1+t)\log(1+t)} dt \\ &=& \log(\log(1+x))-\log(\log(2))+\int_1^x \frac{1}{t(1+t)\log(1+t)} dt \end{eqnarray} Note that the integral $\int_1^\infty \frac{1}{t(1+t)\log(1+t)} dt$ converges to about $0.5715$. Thus we have $$ \int_1^x \frac{1}{t\log(1+t)} dt = \log \log(1+x) -\log \log 2 + \int_1^\infty \frac{1}{t(1+t)\log(1+t)} dt + o(1) $$

With an estimate on $\int_x^\infty \frac{1}{t(1+t)\log(1+t)} dt$ you can get a tighter asymptotic than $o(1)$, for example $$ \lim\limits_{x\rightarrow\infty} \frac{\int_x^\infty \frac{1}{t(1+t)\log(1+t)} dt}{\frac{1}{x\log x}} = 1 $$ gives us $$ \int_1^x \frac{1}{t\log(1+t)} dt = \log \log(1+x) -\log \log 2 + \int_1^\infty \frac{1}{t(1+t)\log(1+t)} dt-\frac1{x\log x} + o\left(\frac1{x\log x}\right) $$

You can proceed likewise to get tighter asymptotics.