What is the integral from i to i/2 of e^{piz} ?

The integral from i to i/2 of e^{piz} is 1/pi (e^{(pii)/2}-e^{pii})

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integral from i to i/2 of e^{piz}

\int_{i}^{\frac{i}{2}} e^{π z} d z

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

Solution steps

\int_{i}^{\frac{i}{2}} e^{π z} d z

Apply u-substitution

: \int_{πi}^{\frac{πi}{2}} e^{u} \frac{1}{π} d u

= \int_{πi}^{\frac{πi}{2}} e^{u} \frac{1}{π} d u

Take the constant out:

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

= \frac{1}{π} \cdot \int_{πi}^{\frac{πi}{2}} e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{π} [e^{u}]_{πi}^{\frac{πi}{2}}

Compute the boundaries:

e^{\frac{πi}{2}} - e^{πi}

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

\int_{0}^{\frac{π}{4}} x sin^{2} (x) d x

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\int_{0}^{3} (- 2 x^{2} + 4 x) d x

\int_{- 1}^{4} (2 - 2 x) d x

\int_{0}^{2} 3 y d y

What is the integral from i to i/2 of e^{piz} ?
The integral from i to i/2 of e^{piz} is 1/pi (e^{(pii)/2}-e^{pii})