Matematické Fórum

stuart clark · 22. 01. 2019 07:21

Finding $\bigg(\int^{\infty}_{-\infty}(-1)^{\frac{1}{1+e^x}}dx\bigg)\cdot \bigg(\int^{\pi}_{0}\frac{\sin x}{x}dx\bigg)^{-1}$

laszky · 24. 01. 2019 19:36

↑ stuart clark:

Hi, how do you define $(-1)^x$ , for $x\in\mathbb{R}\setminus\mathbb{Q}$ ?

stuart clark · 24. 01. 2019 19:38

@↑ laszky: asked by someone and he told me answer is $2i.$ where $i=\sqrt{-1}.$

laszky · 24. 01. 2019 19:43

↑ stuart clark:

So probably $(-1)^x=(\mathrm{e}^{\pi i})^x=\mathrm{e}^{\pi x i}=\cos(\pi x)+i\sin(\pi x)$ .

laszky · 25. 01. 2019 22:33 — Editoval laszky (25. 01. 2019 22:36)

Hi.

stuart clark · 27. 01. 2019 16:30

#1 22. 01. 2019 07:21