Little curiosity $\Re\Big(\int_{0}^{1}\frac{x\ln(x)}{1+e^{ix}}dx\Big)=-\frac{1}{8}$

Question

Playing with the integral of the Glaisher–Kinkelin constant with $x\ln(x)$ I have found a little curiosity :

$$\Re\Big(\int_{0}^{1}\frac{x\ln(x)}{1+e^{ix}}dx\Big)=-\frac{1}{8}$$

I have tried the residue theorem without success .I can't find the right track to apply this. Obviously I have tried to apply the Euler's formula .Moreover the imaginary part looks more complicated that's why I think it's not easy to calculate this directly.I have tried also to differentiate under the integral but it doesn't looks good ...Maybe add a parameter in the exponent of the exponential function ?Maybe we have

$$\Re\Big(\int_{0}^{1}\frac{x\ln(x)}{1+e^{i\alpha x}}dx\Big)=-\frac{1}{8}$$

Where $0<\alpha\leq \pi$ a real number

How to solve it ? is it well-know ?

Any helps is greatly appreciated

Thanks in advance !

Update :

Since it's too easy I add the following question :

What's the value of the parameter $\alpha$ such that :

$$-\Im\Big(\int_{0}^{1}\frac{x\ln(x)}{1+e^{i\alpha x}}dx\Big)=\Re\Big(\int_{0}^{1}\frac{x\ln(x)}{1+e^{i\alpha x}}dx\Big)=-\frac{1}{8}$$

Thanks again!

Answer 1

Note,

$$Re\left(\frac1{1+e^{ix}}\right) = Re\left(\frac{e^{-\frac{ix}{2}}}{e^{\frac{ix}{2}} + e^{-\frac{ix}{2}}}\right) = Re\left(\frac{\cos\frac x2-i \sin \frac x2}{2\cos\frac x2}\right)=\frac12 $$

Thus,

$$\Re\Big(\int_{0}^{1}\frac{x\ln(x)}{1+e^{ix}}dx\Big) =\frac12 \int_{0}^{1}{x\ln(x)}dx $$