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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?

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    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). – 2017-01-26
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    The integral is divergent. (at $x=e^{-1}$ the integrand is logarithmically divergent) – 2017-01-26
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    The integrand is complex for $x < e^{-1}$ – 2017-01-26
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    @mickep Which $x$ is missing ? $x$ of differentiation gets cancelled by $x$ of integration.The question is asked in my homework assignment. – 2017-01-26
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    @Mathgeek, are you sure there is no typo? This is exactly that was in the assignment? – 2017-01-26
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    Ah, sorry. No $x$ is missing. Same change of variables seem good, though. But as others mention, something is perhaps missing/wrong with this exercise. What is the course? – 2017-01-26

2 Answers 2

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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 :

li$(t)=$Ei$(\ln(t))$.

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

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

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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 $1+\log{x}$.