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Integral of $\int_{\Re\lambda=\gamma_0} \frac{e^{\lambda t}}{\lambda^2+c^2}d\lambda$ where $\gamma_0>0$ and $c>0$.

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I want to deform the contour to the imaginary axis but avoiding the poles $\lambda=\pm ic$. To avoid the poles one can use half-circles $C_\pm=\{\lambda\in\mathbb C:\lambda = \pm ic+\epsilon e^{i\theta},-\frac{\pi}{2}\leq\theta\leq\frac{\pi}{2}\}$ with centre $\pm i c$ and of radius $\epsilon$. First I write $$\frac{1}{\lambda^2+c^2}=\frac{1}{2ic}\Big(\frac{1}{\lambda-ic}-\frac{1}{\lambda+ic}\Big).$$ This yields, for $C_+$, $$\int_{C_+}\frac{1}{\lambda^2+c^2}\,d\lambda=\frac{\epsilon}{2c}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}e^{i\theta}e^{(ic+\epsilon e^{i\theta})t}\Big(\frac{1}{(ic+\epsilon e^{i\theta})-ic}-\frac{1}{(ic+\epsilon e^{i\theta})+ic}\Big)\,d\theta$$but I'm not too sure how to proceed.

Is this the best way to compute$$\int_{\Re\lambda=\gamma_0} \frac{e^{\lambda t}}{\lambda^2+c^2}\,d\lambda$$ or is there an easier way?


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