What is the integral from 0 to pi of xe^{-inx} ?

The integral from 0 to pi of xe^{-inx} is ((-1)^n-1}{n^2}+i\frac{pi(-1)^n)/n

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to pi of xe^{-inx}

\int_{0}^{π} x e^{- in x} d x

\frac{( - 1 ) ^{n} - 1}{n ^{2}} + i \frac{π ( - 1 ) ^{n}}{n}

Solution steps

\int_{0}^{π} x e^{- in x} d x

Simplify

x e^{- in x} : x cos (n x) - i x sin (n x)

= \int_{0}^{π} x cos (n x) - i x sin (n x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{π} x cos (n x) d x - \int_{0}^{π} i x sin (n x) d x

\int_{0}^{π} x cos (n x) d x = \frac{( - 1 ) ^{n} - 1}{n ^{2}}

\int_{0}^{π} i x sin (n x) d x = - \frac{πi ( - 1 ) ^{n}}{n}

= \frac{( - 1 ) ^{n} - 1}{n ^{2}} - (- \frac{πi ( - 1 ) ^{n}}{n})

Simplify

\frac{( - 1 ) ^{n} - 1}{n ^{2}} - (- \frac{πi ( - 1 ) ^{n}}{n}) : \frac{( - 1 ) ^{n} - 1}{n ^{2}} + i \frac{π ( - 1 ) ^{n}}{n}

= \frac{( - 1 ) ^{n} - 1}{n ^{2}} + i \frac{π ( - 1 ) ^{n}}{n}

\int_{0}^{t} e^{2 u} d u

\int_{0}^{10} 5 d x

\int_{1}^{2} x^{3} + x d x

\int_{0}^{2 π} 8 sin (θ) cos (θ) d θ

\int_{1}^{7} \frac{5}{x} - \frac{5}{x ^{2}} d x

What is the integral from 0 to pi of xe^{-inx} ?
The integral from 0 to pi of xe^{-inx} is ((-1)^n-1}{n^2}+i\frac{pi(-1)^n)/n