Integration by parts does not work for this complex integral. Why?

Question

There is a longer integral for which integration by parts $\displaystyle\int udv=uv-\int vdu$ was attempted as it came across in research:

$$ \frac i{2\pi}\int_0^{2\pi}\underbrace{\ln\left(1+\frac{e^{-i t}+1}a\ln\left(1-\frac1be^{\frac{e^{it}+1}a}\right)\right)}_u \;d\bigg[\underbrace{\ln\left(e^\frac{e^{it}+1}c-b\right)}_v\bigg] \tag{0} $$

as it has the below problem, but here is a simpler one to better get the point across

$$ \frac1{2\pi}\int_0^{2\pi} \ln(e^{it}+a)\; d\left[\ln\left(e^{e^{it}}-2\right)\right] \tag{1} $$

Integrating $(1)$ by parts should give:
$$\frac1{2\pi}\int_0^{2\pi}\ln(e^{it}+a) \;d\left[\ln\left(e^{e^{it}}-2\right)\right] \\ = \underbrace{\left[\ln(e^{it}+a)\ln\left(e^{e^{it}}-2\right)\middle]\right|_0^{2\pi}}_\text{should equal zero when plugging in directly}-\frac1{2\pi}\int_0^{2\pi}\ln\left(e^{e^{it}}-2\right) \;d\left[\ln(e^{it}+a)\right] \ . \tag2 $$

Directly substituting, we should get $\left[\ln(e^{it}+a)\ln\left(e^{e^{it}}-2\right)\middle]\right|_0^{2\pi} =0$. Checking numerically shows $(2)$ to be false.

However, instead, we numerically get $\ln(a-1)i$ instead of $0$:

$$ \frac1{2\pi}\int_0^{2\pi}\ln(e^{it}+a) \;d\left[\ln\left(e^{e^{it}}-2\right)\right] \\ =\ln(a-1)i -\frac1{2\pi}\int_0^{2\pi}\ln\left(e^{e^{it}}-2\right) d\left[\ln(e^{it}+a)\right] \tag{3} \ . $$

What is the correct way to find the extra $[uv]|_a^b$ term, like $\ln(a-1)i$ here?

Answer 1

According to WolframAlpha, we get $$\lim_{t\to\color{red}{{\pi}^-}}\ln\left(e^{e^{it}}-2\right)=i\pi-1 +\ln(2e-1)$$ and $$\lim_{t\to\color{red}{{\pi}^+}}\ln\left(e^{e^{it}}-2\right)=-i\pi-1+\ln(2e-1)$$

So, if I'm not mistaken, these should give us $(3)$ as follows :

$$\begin{align}&\frac1{2\pi}\int_0^{2\pi}\ln(e^{it}+a) \;d\left[\ln\left(e^{e^{it}}-2\right)\right] \\\\&=\lim_{c_1\to{\pi}^-}\frac1{2\pi}\int_0^{c_1}\ln(e^{it}+a) \;d\left[\ln\left(e^{e^{it}}-2\right)\right]+\frac1{2\pi}\lim_{c_2\to{\pi}^+}\int_{c_2}^{2\pi}\ln(e^{it}+a) \;d\left[\ln\left(e^{e^{it}}-2\right)\right] \\\\&=\lim_{c_1\to {\pi}^-}\bigg(\frac1{2\pi}\left[\ln(e^{it}+a)\ln\left(e^{e^{it}}-2\right)\middle]\right|_0^{c_1}-\frac1{2\pi}\int_0^{c_1}\ln\left(e^{e^{it}}-2\right) \;d\left[\ln(e^{it}+a)\right]\bigg)+\lim_{c_2\to{\pi}^+}\bigg(\frac1{2\pi}\left[\ln(e^{it}+a)\ln\left(e^{e^{it}}-2\right)\middle]\right|_{c_2}^{2\pi}-\frac1{2\pi}\int_{c_2}^{2\pi}\ln\left(e^{e^{it}}-2\right) \;d\left[\ln(e^{it}+a)\right]\bigg) \\\\&=\frac1{2\pi}\bigg(\ln(a-1)\times\bigg(i\pi-1+\ln(2e-1)\bigg)-\ln(a+1)\ln(e-2)\bigg)+\frac1{2\pi}\bigg(\ln(a+1)\ln(e-2)-\ln(a-1)\times \bigg(-i\pi-1+\ln(2e-1)\bigg)\bigg)-\frac1{2\pi}\int_0^{2\pi}\ln\left(e^{e^{it}}-2\right) d\left[\ln(e^{it}+a)\right] \\\\&=\ln(a-1)i -\frac1{2\pi}\int_0^{2\pi}\ln\left(e^{e^{it}}-2\right) d\left[\ln(e^{it}+a)\right]\end{align}$$

Answer 2

Issue :

When we write
$$ \frac1{2\pi}\int_0^{2\pi} x\ln(e^{it}+a)\; dx \tag{X} $$ we are saying that the limits are for $x$ (it is not for $a$ , not for $t$)

Likewise , when we write
$$ \frac1{2\pi}\int_0^{2\pi} \ln(e^{it}+a)\; d\left[\ln\left(e^{e^{it}}-2\right)\right] \tag{Y} $$ we are telling Wolfram that the limits are for $\left[\ln\left(e^{e^{it}}-2\right)\right]$ (it is not for $a$ , not for $t$)

Hence we have given the limits to Wolfram :
$\left[\ln\left(e^{e^{it}}-2\right)\right]=0$
$\left[\ln\left(e^{e^{it}}-2\right)\right]=2\pi$

Over-all , the limits which will Wolfram use for $t$ are :
$t=\ln\ln3/i$
$t=\ln\ln[e^{2\pi}+2]/i$

This is not a closed curve.
When we plug that into (2) , naturally , we will not get $0$ where we might want it.

ADDENDUM :

When we Integrate By Parts to get (3) with the new $d[[X]]$ term involving $\left[\ln\left(e^{e^{it}}+a\right)\right]$ , we have to handle the limits correctly & consistently for the "change of variable" there.
When we make the limit changes through-out , the extra term should work out to what Wolfram gave.

Now , the limits Wolfram will see are :
$\left[\ln\left(e^{e^{it}}+a\right)\right]=0$
$\left[\ln\left(e^{e^{it}}+a\right)\right]=2\pi$

Hence , the limits Wolfram will use are :
$t=\ln\ln(1-a)/i$
$t=\ln\ln(e^{2\pi}+a)/i$

SUMMARY :

The Crux is that manually checking gave a $0$ which is wrong because the limits were wrong.
Essentially , the manual work is involving 2 unrelated Integrals.

Difference between those 2 Integrals is coincidentally $\log(a-1)i$ , which Wolfram correctly calculates.