Can the Fourier integral of sinw be solved with a misprint?

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I'm a little unsure of how the attached solutions are obtained. For instance one of the Fourier integrals is:

∫e(wE-iw)t = [1/(wE-iw)e(wE-iw)t from -∞ to ∞. But what do you do about the real part wE? That leads the integral to diverge as far as I can see...
 

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hizezima1! :smile:

those ewEts are a misprint, they should be eiwEt :wink:
 
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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