Evaluate the following

$$\int_{0}^{\pi/4} \ln(1+\ln(x))\cdot dx$$

I was trying to do it using by parts but got stuck at $\int \frac{1}{1+\ln(x)} dx$

Could someone suggest how to proceed fro here or any better method?

  • 4
    $\begingroup$ The integral looks a bit unnatural, $\pi/4$ is an uncommon argument of the logarithm. Do you mind me asking for the source of this integral? (There is an $x$ missing in the integral you got after integrating by parts. If you do $u=1+\ln x$ in that integral you will encounter an exponential integral). $\endgroup$ – mickep Jan 26 '17 at 8:58
  • $\begingroup$ The integral is divergent. (at $x=e^{-1}$ the integrand is logarithmically divergent) $\endgroup$ – Fabian Jan 26 '17 at 9:07
  • $\begingroup$ The integrand is complex for $x < e^{-1}$ $\endgroup$ – gammatester Jan 26 '17 at 9:08
  • $\begingroup$ @mickep Which $x$ is missing ? $x$ of differentiation gets cancelled by $x$ of integration.The question is asked in my homework assignment. $\endgroup$ – Mathgeek Jan 26 '17 at 9:09
  • $\begingroup$ @Mathgeek, are you sure there is no typo? This is exactly that was in the assignment? $\endgroup$ – Yuriy S Jan 26 '17 at 9:51

This is not an answer, but a comment too long to be put in the comments section.

As already pointed out, there is probably a typo in the wording since $\ln(1+\ln(x))$ isn't real in $x\leq e^{-1}$.

Considering the indefinite integral, the integration by part leads to:

$$\int \ln(1+\ln(x)) dx=x\ln(1+\ln(x))-\int \frac{dx}{1+\ln(x)}$$ $\int \frac{dx}{1+\ln(x)}=\frac{1}{e}\int \frac{d(ex)}{\ln(ex)}=\frac{1}{e}\int \frac{dt}{\ln(t)}=\frac{1}{e}\text{li}(t)+c=\frac{1}{e}\text{li}(ex)+c$

li$(t)$ is the logarithmic integral function which is related to the Exponential integral function :


$\int \frac{dx}{1+\ln(x)}=\frac{1}{e}\text{Ei}(\ln(ex))+c=\frac{1}{e}\text{Ei}(1+\ln(x))+c$

$$\int \ln(1+\ln(x)) dx=x\ln(1+\ln(x))- \frac{1}{e}\text{Ei}(1+\ln(x))+c$$ The result is not real if the lower bound of the integral is $\leq e^{-1}$.


Indefinite integral is $-\frac{-\int_{-(1+\log{(x)})}^{\infty}\exp{(-t)}t^{-1}dt}{\exp{x}}+x\log{(1+\log{(x)})}$, where the term in the numerator is exponential integral function at $1+\log{x}$.


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