# Tag Info

## Hot answers tagged integration

5

This is true for $a\in \mathbb Z$ too. $-(\pi-x)a$ is real, so $-i(\pi-x)a$ is purely imaginary. It is well known that $$e^{it}=\cos t + i\sin t \qquad\qquad\text{when }t\in\mathbb R$$ so its modulus is $\sqrt{\sin^2 t+\cos^2 t} = \sqrt 1 = 1$.

4

The image below might help to explain where this term comes from:

4

We have: $$I=\int_{0}^{\pi/2}e^{-\sin x}\,dx = \int_{0}^{1}\frac{e^{-t}}{\sqrt{1-t^2}}\,dt$$ and since: $$e^{-t}=\sum_{k\geq 0}\frac{(-1)^k t^k}{k!},\qquad \int_{0}^{1}\frac{t^k}{\sqrt{1-t^2}}\,dt =\int_{0}^{\pi/2}\sin^k\theta\,d\theta=\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{k+1}{2}\right)}{2\,\Gamma\left(\frac{k}{2}+1\right)}$$ it follows ...

3

Depict carefully the path of integration: it is a semicircle in the upper half plane with a bulge at $z=0$ and a keyhole around $z=i$. This gives that you have to compute the residues of $f(z)=\frac{e^{iz}}{z(z^2+1)^2}$ at $z=0$ and $z=i$, but to consider only half the residue at $z=0$: ...

3


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