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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

jjagmath
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MeltedStatementRecognizing
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    You should not do $x = u * \sqrt{i}$. For solving it in a complex analysis way, maybe see: https://math.stackexchange.com/questions/1100339/integrate-int-0-infty-fracdxx22x122-using-residues – onRiv May 17 '22 at 14:54
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    You can try to do $x = \frac{1}{t}$. It reduces to $\int_{0}^{\infty} \frac{t^2}{1+t^4} \text{d} x$, hence $\frac{1}{2}\int_{0}^{\infty}\frac{1+t^2}{1+t^4}dt$. Then do $t-\frac{1}{t}=u$. – onRiv May 17 '22 at 15:11
  • Avoid the use of $*$ to denote multiplication. That's a common practice in programming, not in Mathematics. – jjagmath May 17 '22 at 15:22

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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 $\{ \frac{u}{\sqrt i} \text{ }| \text{ } u \ge 0 \}.$

Asem Abdelraouf
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