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How can I evaluate this integral $$\int_0^{+\infty}\frac{e^{-r}}{1+r^2}dr$$ ?

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1 Answer 1

up vote 4 down vote accepted

You can evaluate it in terms of the sine and cosine integrals.

$ \displaystyle \int^{\infty}_{0} \frac{e^{-x}}{1+x^{2}} \ dx = \int_{0}^{\infty}\int_{0}^{\infty} e^{-x} \sin t \ e^{-xt} \ dt \ dx$

$ \displaystyle = \int^{\infty}_{0} \int_{0}^{\infty} \sin t \ e^{-(t+1)x} \ dx \ dt $ since the iterated integral converges absolutely

$\displaystyle = \int_{0}^{\infty} \frac{\sin t}{t+1} \ dt = \int_{1}^{\infty} \frac{\sin (u-1)}{u} \ du $

$ \displaystyle= \cos 1 \int_{1}^{\infty} \frac{\sin u}{u} \ du - \sin 1 \int_{1}^{\infty} \frac{\cos u}{u} \ du $

$ \displaystyle = \cos 1 \Big(\int_{0}^{\infty} \frac{\sin u}{u} \ du - \int_{0}^{1} \frac{\sin u}{u} \ du \Big) + \sin 1 \ \text{Ci}(1) $

$ \displaystyle = \cos 1 \Big(\frac{\pi}{2} - \text{Si}(1) \Big) + \sin 1 \ \text{Ci} (1) $

$ \displaystyle= \frac{\pi}{2} \cos 1 - \cos 1 \ \text{Si} (1) + \sin 1 \ \text{Ci} (1) $

EDIT:

If you're just interested in $\displaystyle \lim_{\epsilon \to 0^{+}} \int_{0}^{\infty}\frac{e^{-\epsilon x}}{1+x^{2}} \ dx$, it's much easier to just justify brining the limit inside of the integral.

But $\displaystyle \displaystyle \lim_{\epsilon \to 0^{+}} \int_{0}^{\infty}\frac{e^{-\epsilon x}}{1+x^{2}} \ dx = \lim_{\epsilon \to 0^{+}} \Big( \frac{\pi}{2} \cos \epsilon - \cos \epsilon \ \text{Si} (\epsilon) + \sin \epsilon \ \text{Ci} (\epsilon) \Big)$

$ \displaystyle= \frac{\pi}{2} - 0 + \lim_{\epsilon \to 0^{+}} \sin \epsilon \ \text{Ci} (\epsilon) = \frac{\pi}{2} + \lim_{\epsilon \to 0^{+}} \sin \epsilon \left(\ln \epsilon + \gamma + O(\epsilon^{2}) \right) = \frac{\pi}{2}$

SECOND EDIT:

$\displaystyle \displaystyle \lim_{\epsilon \to \infty} \int_{0}^{\infty}\frac{e^{-\epsilon x}}{1+x^{2}} \ dx = \lim_{\epsilon \to \infty} \Bigg( \cos \epsilon \Big( \frac{\pi}{2} - \text{Si}(\epsilon) \Big) + \sin \epsilon \ \text{Ci} (\epsilon) \Bigg)$

$\displaystyle = 0 + \lim_{\epsilon \to \infty} \sin \epsilon \ \text{Ci}(\epsilon)$

For large $\epsilon$, $\text{Ci}(\epsilon) \approx \frac{\sin \epsilon}{\epsilon} $.

So the above limit is zero.

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What $Si$ and $Ci$? –  Sue Apr 21 '13 at 22:22
    
    
Maybe I can solve easily; I'm really interested with $$\lim_{\varepsilon\to 0}\int_0^{+\infty}\frac{e^{-\varepsilon r}}{1+r^2}$$ so I can use the theorem of dominated convergence –  Sue Apr 21 '13 at 22:38
    
If I have $$\int_{0}^{+\infty}\frac{e^{-\varepsilon r}}{1+r^2}$$ what result I obtain with your reasoning? –  Sue Apr 21 '13 at 22:55
    
And if $\varepsilon\to+\infty$? With dominated convergence the result is $0$. With your reasoning? –  Sue Apr 22 '13 at 11:59

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