z-0-t-i-1-x-dx-0-t-ie-ipix-dx-1-i-0-t-e-ipix-dx-i-1-ipi-e-ipit-1-z-1-pi-1-t-1-2-0-t-ie-ipix-dx-u-e-ipix-du-1-ipi-e-ipix-dx-0-t-ipi-i

Question Number 7021 by FilupSmith last updated on 06/Aug/16

z = \int_{0}^{t} i {(- 1)}^{x} d x

= \int_{0}^{t} {i e}^{i π x} d x

(1)

= i \int_{0}^{t} e^{i π x} d x

= i (\frac{1}{i π} (e^{i π t} - 1))

z = \frac{1}{π} ({(- 1)}^{t} - 1)

(2)

= \int_{0}^{t} {i e}^{i π x} d x

u = e^{i π x} \Rightarrow d u = \frac{1}{i π} e^{i π x} d x

= \int_{0}^{t} \frac{i π}{i π} {i e}^{i π x} d x

= \int_{0}^{t} i π i d u

= - π \int_{0}^{t} d u

= - π (e^{i π t} - 1)

which is incorrect ?

Commented by FilupSmith last updated on 07/Aug/16

Ah, i see my mistake, thank you!

Commented by Yozzii last updated on 06/Aug/16

I n (2), f o r u = e^{i π x} \Rightarrow d u = i π e^{i π x} d x

\Rightarrow u = e^{π i t} a t x = t a n d u = 1 a t x = 0 .

\Rightarrow \int_{0}^{t} {i e}^{π i x} d x = i \times \frac{1}{i π} \int_{1}^{e^{i π t}} d u = \frac{1}{π} (e^{π i t} - 1) = \frac{1}{π} ({(- 1)}^{t} - 1)

Answered by Yozzii last updated on 08/Aug/16

C h e c k f o r a r e s p o n s e i n t h e c o m m e n t s .

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