Integral Help: \int arg(x^{ix} + x^{-ix}) dx

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I really have no idea how to do this, please show some steps.

\int arg(x^{ix} + x^{-ix}) dx
 
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If x is real the exp(ix)+exp(-ix) is real. Why? What is arg of a real number? Are you sure that's the whole problem?
 
Dick said:
If x is real the exp(ix)+exp(-ix) is real. Why? What is arg of a real number? Are you sure that's the whole problem?

I'm pretty sure that when x is real, the inside does not necessarily equal a real number. I'm looking at the graph right now and y = arg(...) is not zero, and also this isn't homework.
 
Hmm. Ok, what is x? Do they give you a contour to integrate along? It's pretty tricky to define the general complex power of a complex number.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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