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Some fellow students were talking in a room a while back and apparently they're calculus professor told them a random integral they wrote up was "unsolvable" at the calculus semester 2 level.

The integral was $x \tan(x)$.

To try and see if I could solve it for them (out of curiosity) I was able to do the following by the method of integration of parts:

$\int x \tan(x) dx = x \int \tan(x) dx - \int \int \tan(x) dxdx$

Then by plugging in the integral of tangent:

$-x\ln|\cos(x)| + \int \ln|\cos(x)|dx$

The absolute value of cos can be rewritten as the absolute value of sin which can be rewritten via modulo:

$-x\ln|\cos(x)| + \int \ln|\sin(x+\frac \pi2)|dx = -x\ln|\cos(x)| + \int \ln\sin((x - \frac \pi2) \mod \pi) dx$

I can completely substitute away the modulo operation as I know how to adjust for such a substitution in the general case. (I presumed that this was the issue the professor referred to as most students do not learn of such functions). That leaves me with:

$-x\ln|\cos(x)| + \int \ln\sin(u)du$

This gets me to the final issue I cannot seem to solve. What is the integral of ln(sin(x))? I hear it has no closed form, yet when me and the other students looked it up, it said something about "poly-logarithms"? Is that some kind of made up function used to define an integral with no closed form? What does it mean?

user64742
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    The integral in question can be written in terms of the dilogarithm function $\text{Li}_2(z)$, which can be expressed as

    $$\text{Li}_2(z)=-\int_0^z \frac{\log(1-z')}{z'},dz'$$

    for $z\in \mathbb{C}-[1,\infty)$.

     – Mark Viola Jun 28 '16 at 02:43
  • Obviously, if the definition holds for complex values, it holds for those complex numbers with zero imaginary parts. But, one may write the answer using complex components. – Mark Viola Jun 28 '16 at 02:48
  • Closely related (duplicate?): http://math.stackexchange.com/questions/740911/what-is-int-x-tanxdx – Martin Sleziak Jun 28 '16 at 18:51
  • One more question related to integral of $x\tan x$: http://math.stackexchange.com/questions/847829/int-xtanx-and-the-clausen-function Since you ask also about $\ln\sin x$: http://math.stackexchange.com/questions/356543/integral-of-ln-sinx and http://math.stackexchange.com/questions/794916/today-a-student-asked-me-int-ln-sin-x-dx – Martin Sleziak Jul 01 '16 at 04:32
  • Please, try to make the titles of your questions more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Martin Sleziak Jul 01 '16 at 05:05
  • I don't know that this would work, but have you tried a series expansion, and perhaps just integrating the series expansion? That would qualify as a solution, I would think. – johnnyb May 03 '23 at 13:15

1 Answers1

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Let $I$ be the indefinite integral given

$$I=\int x\tan(x)\,dx \tag 1$$

Integrating by parts the integral in $(1)$ with $u=x$ and $v=\log(\cos(x))$ reveals

$$\begin{align} I&=-x\log(\cos(x))+\int \log(\cos(x))\,dx \\\\ &=-x\log(\cos(x))+\int \log\left(\frac{e^{ix}+e^{-ix}}{2}\right)\,dx \\\\ &=-x\log(\cos(x))-\log(2)x+\int \log\left(e^{ix}+e^{-ix}\right)\,dx\\\\ &=-x\log(\cos(x))-\log(2)x-\frac i2 x^2 +\int \log\left(1+e^{i2x}\right)\,dx\tag 2 \end{align}$$

Now, enforcing the substitution $u=-e^{i2x}$ in the integral of $(2)$ reveals

$$\begin{align} \int \log(1+e^{i2x})\,dx&=\int \frac{\log(1-u)}{i2u}\,du\\\\ &= \frac i2 \text{Li}_2(-e^{i2x}) \tag 3 \end{align}$$

Substituting $(3)$ in $(2)$ yields

$$\begin{align} I&=-x\log(\cos(x))-\log(2)x-\frac i2 x^2+\frac i2 \text{Li}_2(-e^{i2x})+C\\\\ &=\bbox[5px,border:2px solid #C0A000]{-x\log\left(1+e^{i2x}\right)+\frac i2 x^2+\frac i2 \text{Li}_2(-e^{i2x})+C} \end{align}$$

Mark Viola
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