I'm trying to integrate the following:

$$I = \int^\infty_0 \dfrac{\cos x}{x^2 + 1} dx$$

What I did was:

$$I = \int^\infty_0 \dfrac{\cos x}{x^2 + 1} dx = \dfrac{1}{2}\int^\infty_{-\infty} \dfrac{e^{ix}+e^{-ix}}{2(x^2+1)} dx = \dfrac{1}{4}\int^\infty_{-\infty} \dfrac{e^{ix}}{x^2+1} + \dfrac{1}{4}\int^\infty_{-\infty} \dfrac{e^{-ix}}{x^2+1} dx$$

Now I'm not sure what to do. I'm aware that I'm supposed to solve this using Jordan's Lemma (I think?), though I don't know how.

Let's call $$I_1 = \dfrac{1}{4}\int^\infty_{-\infty} \dfrac{e^{ix}}{x^2+1}$$ and

$$I_2 = \dfrac{1}{4}\int^\infty_{-\infty} \dfrac{e^{-ix}}{x^2+1} dx$$

I know that both $I_1$ and $I_2$ have simple poles in $Z = \pm i$. $I_1$ is easy to calculate, as:

$$I_1 = 2\pi i R_{z=i}$$, where,

$$R_{z=i} = lim_{z-> i} (z-i)\dfrac{e^{iz}}{(z -i)(z+i)} = \dfrac{1}{2ei}$$

Now, what should I do with $I_2$? Can I do exactly the same? Why or why not?


Note that making the change of variables $y=-x$ in $I_2$ you get $I_2=I_1$.

  • 1
    $\begingroup$ What you're saying is that $I = 2I_1$? $\endgroup$ – iamatrain Jan 27 '16 at 22:28
  • 1
    $\begingroup$ Precisely. ---- $\endgroup$ – John B Jan 27 '16 at 22:29

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