$$\int_{-\infty}^{\infty} \frac{\cos x}{1+x^2} e^{-ixt} \,\mathrm dx \quad \quad \quad \text{for }t>0$$

Use residue formula, which contour should I try?

  • $\begingroup$ I'm not sure, but did you try a semicircle and then use Jordan's lemma? $\endgroup$
    – MEDVIS
    Dec 5, 2014 at 12:36

1 Answer 1


$$\int_{-\infty}^{\infty} \frac{\cos x}{1+x^2} e^{-ixt} \,\mathrm dx=\int_{-\infty}^{\infty} \frac{\cos(x)\cos(xt)}{1+x^2} \,\mathrm dx$$ because $$Im\int_{-\infty}^{\infty} \frac{\cos x}{1+x^2} e^{-ixt} \,\mathrm dx=0$$

now because $$\cos(x)\cos(xt)=\frac{cos(x(1+t))+cos(x(1-t))}{2}$$ Use this Contour

enter image description here

to see that get $$\int_{-\infty}^{\infty} \frac{e^{ixt}}{1+x^2} \,\mathrm dx \,\mathrm =\int_{-\infty}^{\infty} \frac{cos(xt)}{1+x^2} \,\mathrm dx \,\mathrm=\pi\ e^{-|t|} $$

and so $$\int_{-\infty}^{\infty} \frac{\cos(x)\cos(xt)}{1+x^2} \,\mathrm dx=\pi\frac{e^{-|t+1|}+e^{-|t-1|}}{2}$$ finally

$$\int_{-\infty}^{\infty} \frac{\cos x}{1+x^2} e^{-ixt} \,\mathrm dx=\frac\pi2\left({e^{-|t+1|}+e^{-|t-1|}}\right)$$


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