2
  1. $\int_0^\infty \frac{\cos (x)}{1+4x^2}\, dx$
  2. $\int_0^\infty \frac{1}{(1+x^2)^2}\, dx$

There is no hint for these two questions. I think for Q2, since it's a square, I can use Plancherel formula for $e^{-2\pi|x|}$. But I am not sure how to solve the first one.

John
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  • Hint : from the complex expression of $\cos(x)$ rewrite 1. as a Fourier integral over $(-\infty,+\infty)$. – Raymond Manzoni Nov 29 '14 at 15:49
  • @RaymondManzoni Then view $\frac{1}{1+x^2}$ as Fourier transform of some functions and use multiplicative formula? – John Nov 29 '14 at 16:03
  • I spoke only of $1.$ for which you obtain the Fourier transform of $\dfrac 1{1+4x^2}$ at point $1$ (with a coefficient depending of your definition of the Fourier transform). – Raymond Manzoni Nov 29 '14 at 16:37

1 Answers1

2

Let $f(t) = e^{-|t|}$, then with $\hat{f} = {\cal F} f$, we have $\hat{f}(x) ={2 \over 1+x^2}$. The inversion theorm gives

\begin{eqnarray} f(t) &=& {1 \over 2\pi}\int_{\mathbb{R}} {2 e^{i xt} \over 1 + x^2} dx\\ &=& {1 \over \pi}\int_{-\infty}^0 {e^{i xt} \over 1 + x^2} dx + {1 \over \pi}\int_0^\infty {e^{i xt} \over 1 + x^2} dx\\ &=& {1 \over \pi}\int_0^\infty {e^{i xt} + e^{-ixt}\over 1 + x^2} dx\\ &=& {2 \over \pi} \int_0^\infty {\cos(xt)\over 1 + x^2} dx \\ &=& {4 \over \pi} \int_0^\infty {\cos(2yt)\over 1 + (2y)^2} dy \\ \end{eqnarray} and so (1) is given by ${\pi \over 4} f({1 \over 2})$.

Look at $f(0)$ and the Plancherel/Parseval identity for (2).

copper.hat
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  • Even though my definition of Fourier transform is different from yours (I follow the book by Stein), I think I understand the approach. Thanks. – John Nov 29 '14 at 17:08
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    Some authors have a ${1 \over \sqrt{2 \pi}}$ factor in the transform to 'symmetrise' the 'forward' and 'reverse' transforms. – copper.hat Nov 29 '14 at 18:01