1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 $z\in \mathbb{C}-[1,\infty)$. $\endgroup$ – Mark Viola Jun 28 '16 at 2:43
  • $\begingroup$ @Dr.MV as far as I was aware, this was a real valued functiom integration. Does some transition to complex numbers take place? That would make sense I suppose. I hear that particular professor tends to teach complex number stuff in basic calculus. $\endgroup$ – The Great Duck Jun 28 '16 at 2:45
  • $\begingroup$ 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. $\endgroup$ – Mark Viola Jun 28 '16 at 2:48
  • $\begingroup$ @Dr.MV oh yeah, you're right. I dont know what I was thinking. XD $\endgroup$ – The Great Duck Jun 28 '16 at 2:49
  • $\begingroup$ Closely related (duplicate?): math.stackexchange.com/questions/740911/what-is-int-x-tanxdx $\endgroup$ – Martin Sleziak Jun 28 '16 at 18:51
1
$\begingroup$

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

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

$\endgroup$
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Daniel Fischer Jun 29 '16 at 11:23
  • $\begingroup$ @Dr.MV it's not a "best" vote. It's an answered vote. No. The revoke was in error. Wrong post. $\endgroup$ – The Great Duck Jan 19 '17 at 3:56
  • $\begingroup$ Thank you, much appreciative. $\endgroup$ – Mark Viola Jan 19 '17 at 4:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.