integral from 0 to i of sinh(piz)

Lösung

\int_{0}^{i} sinh (π z) d z

\frac{e ^{- πi} + e ^{πi} - 2}{2 π}

Schritte zur Lösung

\int_{0}^{i} sinh (π z) d z

Umschreiben mit Hilfe von Trigonometrie-Identitäten

= \int_{0}^{i} \frac{- e ^{- π z} + e ^{π z}}{2} d z

Entferne die Konstante:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int_{0}^{i} - e^{- π z} + e^{π z} d z

Wende die Summenregel an:

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

= \frac{1}{2} (- \int_{0}^{i} e^{- π z} d z + \int_{0}^{i} e^{π z} d z)

\int_{0}^{i} e^{- π z} d z = \frac{- e ^{- πi} + 1}{π}

\int_{0}^{i} e^{π z} d z = \frac{1}{π} (e^{πi} - 1)

= \frac{1}{2} (- \frac{- e ^{- πi} + 1}{π} + \frac{1}{π} (e^{πi} - 1))

Vereinfache

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

= \frac{e ^{- πi} + e ^{πi} - 2}{2 π}

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