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Why can't the function $\mathbb{Li}(x)$ be directly evaluated?

I tried using the substitution $e^u=x$, giving $\ln(x)=u$, and then $dx=e^udu$, then using the Reverse Product Rule on it infinitely, but it turns out that that sequence can't be evaluated either.

Any ideas?

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    It *can* be evaluated. I think you mean to say that it can't be written in closed from using elementary functions. – 2017-02-02
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    There's not really a deep reason for it, as far as I know. Some functions just don't have elementary antiderivatives, and $f(x)=\frac{1}{\ln(x)}$ happens to be one of them. Since this $f(x)$ is - apart from some factors - asymptotic to the density of primes as $x\rightarrow\infty$, its integral comes up often enough to be given its own name. – 2017-02-02
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    Another example is the function $e^{-x^2}$ , ironically the most important function in probability theory. – 2017-02-02
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    You might be interested in how the [Risch algorithm](https://en.wikipedia.org/wiki/Risch_algorithm) applies to this particular function, as it "decides" which rational expressions in the elementary transcendental functions have antiderivatives expressed in this same way. Not all such functions do, of course. – 2017-02-02
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    @Peter Yes, that one is a rather remarkable function and it can't be evaluated as well. – 2017-02-03
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    @Winther Ah, yep. That's what I meant. – 2017-02-03
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    Taking anti-derivatives tends to produce new "special functions". E.g. $\int (1/x)dx =\log x$, which is not expressible in closed form using rational functions. – 2017-02-05

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With your way you don't really get a closed form.

You obtain what is called a Series solution. Like for the integral of $x^x$.

There is no primitive of that integral. And you can prove it via Risch algorithm.

https://en.wikipedia.org/wiki/Risch_algorithm

https://en.wikipedia.org/wiki/Nonelementary_integral