I did $ x = u\sqrt{i}$
$$\sqrt{i}\int_{0}^{\infty} \frac{1}{1 - u^4} \, du$$ $$\sqrt{i}\int_{0}^{\infty} \frac{1}{1 - u^2} \cdot \frac{1}{1 + u^2} \, du$$ $ v = \tan^{-1}(u)$,$dv = \frac{1}{1 + u^2} du$
$$\sqrt{i}\int_{0}^{\pi/2} \frac{1}{1 - \tan^2(v)} dv$$ Using King's property, $\sqrt{i}\int_{0}^{\pi/2} \frac{1}{1 - \cot^2(v)} dv$. If I is the integral, 2I = $\sqrt{i}\int_{0}^{\pi/2} \frac{1}{1 - \tan^2(v)} dv$ + $\sqrt{i}\int_{0}^{\pi/2} \frac{1}{1 - \frac{1}{\tan^2(v)}} dv$. $$\sqrt{i}\int_{0}^{\pi/2} \frac{1}{1 - \tan^2(v)} dv + \sqrt{i}\int_{0}^{\pi/2} \frac{-\tan^2(v)}{1 - \tan^2(v)} dv$$ $\sqrt{i}\int_{0}^{\pi/2} 1 dv$ = $\sqrt{i} \cdot \pi/2$. By de Moivre's theorem $(\cos(\pi/2) + i\sin(\pi/2))^\frac{1}{2}$ = $\cos(\pi/4) + i\sin(\pi/4) $
so the answer would be $\frac{\pi}{2\sqrt{2}} + i\frac{\pi}{2\sqrt{2}}$ But Desmos says it's real What is wrong with the way I did it? Thank you
You need to think of the first integral $$\sqrt{i}\int_{0}^{\infty} \frac{1}{1 - u^4} \, du$$ as being an integral of a function defined on the complex plane, and the path of integration is along the half line $\{ \frac{u}{\sqrt i} \text{ }| \text{ } u \ge 0 \}.$