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Can anyone help me prove this with the help of the Laplace transformation?

$$\int_0^\infty \frac{w}{1+w^2} \sin wx dw=\frac{\pi}{2}e^{-x}$$

where $x>0$

EDIT: So I was wondering if you could split up $\sin wx$ into $$\frac{1}{2i}\big[e^{iwx}-e^{-iwx}\big]$$

Then say

$$\int_0^\infty e^{iwx}-e^{-iwx}=\int_{-\infty}^0 e^{-iwx}-\int_0^\infty e^{-iwx}=\int_{-\infty}^\infty -\operatorname{sgn}(x) e^{-iwx}$$

And use this to calculate the above problem? (by using a Fourier transformation)

EDIT#2: ok, so if you define $f(t)=\operatorname{sgn}(x)e^{-|x|} \Rightarrow \hat{f}(w)=-2i\frac{w}{1+w^2}$

So if

$$\int_0^\infty \frac{w}{1+w^2} \sin wx dw=\frac{1}{2i}\int_{-\infty}^\infty \frac{w}{1+w^2}e^{iwx}dw=\frac{\pi}{2} \frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(w)e^{iwx}dw=\frac{\pi}{2}f(x)$$

Then for $x>0$ this gives

$$\int_0^\infty \frac{w}{1+w^2} \sin wx dw=\frac{\pi}{2}e^{-x}$$


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up vote 1 down vote accepted

Define $f(t)=sgn(x)e^{-|x|} \Rightarrow \hat{f}(w)=-2i\frac{w}{1+w^2}$

So if

$$\int_0^\infty \frac{w}{1+w^2} \operatorname{sin}wx dw=\frac{1}{2i}\int_{-\infty}^\infty \frac{w}{1+w^2}e^{iwx}dw=\frac{\pi}{2} \frac{1}{2\pi}\int_{-\infty}^\infty \hat{f}(w)e^{iwx}dw=\frac{\pi}{2}f(x)$$

Then for $x>0$ this gives

$$\int_0^\infty \frac{w}{1+w^2} \operatorname{sin}wx dw=\frac{\pi}{2}e^{-x}$$

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