Complex Contour Integral $\int_C e^{5z-1}$

Question

I am working with complex integrals and the following integral has been causing me difficulties.

$$\int_C e^{5z-1}$$

I want to integrate this around the semicircle $C$ which is $e^{it}$ where $t \in [\frac {-\pi} 2, \frac \pi 2]$

Ok so do I parameterize the semi-circle?

My attempt

If we let $f(z)= e^{5z-1}$ and $\gamma(t)=e^{it}$ and $\gamma'(t)=ie^{it}$

Then $I = \int_{\frac {-\pi} 2} ^ {\frac \pi 2} e^{5e^{it}-1}*ie^{it} dt$

Then $I = \int_{\frac {-\pi} 2} ^ {\frac \pi 2} ie^{6e^{it}-1}dt$

I feel like this method isn't correct. Is there an easier way of approaching this question?? Any help is appreciated!

Answer 1

Let $f(z)=\frac15e^{5z-1}$. Then $f'(z)=e^{5z-1}$ and therefore\begin{align}\int_Ce^{5z-1}\,\mathrm dz&=f(e^{\pi i/2})-f(e^{-\pi i/2})\\&=f(i)-f(-i)\\&=\frac15\left(e^{5i-1}-e^{-5i-1}\right).\end{align}

Answer 2

Alternate method: The integrand is entire, so by Cauchy's theorem the integral around the closed semicircle is zero.

The integral along the imaginary axis is $\int _{-1}^1e^{5yi-1}i\operatorname dy=i[e^{5yi-1}/(5i)]_{-1}^1=1/5(e^{5i-1}-e^{-5i-1})$.

Therefore the integral around the semicircle is $1/5(e^{5i-1}-e^{-5i-1})$.