3
$\begingroup$

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

Use residue formula, which contour should I try?

$\endgroup$
1
  • $\begingroup$ 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

4
$\begingroup$

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

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

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.