integral from-pi to pi of xsin(kx)

解

\int_{- π}^{π} x sin (k x) d x

- \frac{2 π cos ( πk )}{k} + \frac{2 sin ( πk )}{k ^{2}}

解答ステップ

\int_{- π}^{π} x sin (k x) d x

部分積分を適用する

= [\frac{1}{k} (- x cos (k x) - k \cdot \int - \frac{1}{k} cos (k x) d x)]_{- π}^{π}

\int - \frac{1}{k} cos (k x) d x = - \frac{1}{k ^{2}} sin (k x)

= [\frac{1}{k} (- x cos (k x) - k (- \frac{1}{k ^{2}} sin (k x)))]_{- π}^{π}

簡素化

[\frac{1}{k} (- x cos (k x) - k (- \frac{1}{k ^{2}} sin (k x)))]_{- π}^{π} : [\frac{1}{k} (- x cos (k x) + \frac{1}{k} sin (k x))]_{- π}^{π}

= [\frac{1}{k} (- x cos (k x) + \frac{1}{k} sin (k x))]_{- π}^{π}

限度を計算する:

- 2 \cdot \frac{π}{k} cos (πk) + \frac{2}{k ^{2}} sin (πk)

= - 2 \cdot \frac{π}{k} cos (πk) + \frac{2}{k ^{2}} sin (πk)

簡素化

= - \frac{2 π cos ( πk )}{k} + \frac{2 sin ( πk )}{k ^{2}}