After-looking-at-a-previous-question-I-was-wondering-if-the-following-is-correct-I-n-0-n-1-x-dx-n-R-I-n-0-n-e-ipix-dx-1-I-n-i-pi-e-ipix-0-n-I-n-i-pi-e-ipix-0-n-

Question Number 6441 by Temp last updated on 27/Jun/16

After looking at a previous question

I was wondering if the following

is correct :

I (n) = \int_{0}^{n} {(- 1)}^{x} d x, n \in R

I (n) = \int_{0}^{n} e^{i π x} d x (1)

I (n) = - \frac{i}{π} {(e^{i π x})}_{0}^{n}

I (n) = - \frac{i}{π} {(e^{i π x})}_{0}^{n}

I (n) = - \frac{i}{π} (e^{i π n} - 1)

I (n) = \int_{0}^{n} (\cos (x) + i \sin (x)) d x (2)

I (n) = {(\sin (x) - i \cos (x))}_{0}^{n}

I (n) = - i {(\cos (x) + i \sin (x))}_{0}^{n}

I (n) = - i (e^{i π n} - 1)

Which is correct ?

Commented by Yozzii last updated on 27/Jun/16

I n (2) c o s x + i s i n x = e^{i x} \neq e^{π i x} .

\int_{0}^{n} e^{i x} d x = \frac{1}{i} e^{i x} ∣_{0}^{n} = - i (e^{n i} - 1)

(1) i s c o r r e c t, (2) i s i n c o r r e c t .

Commented by Temp last updated on 27/Jun/16

A h h h, thanks! Silly error

Commented by Temp last updated on 27/Jun/16

I checked and if i fix (2) it has the

same result!

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