Convolution of two identical functions and Fourier transform

Question

I have the following problem to solve:

Express the integral $$\int_{-\infty}^\infty \frac{1}{1+x^2}\frac{1}{1+(x-t)^2}dx$$ in terms of $t$.

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My solution is as follows:

We quickly realize that we have that the integral is just the convolution $f * f$ where

$$f := \frac{1}{1+x^2}$$

Furthermore, if we define the integral as $I(t)$, we have that $\mathcal{F}(I) = \mathcal{F}(f * f) = (\mathcal{F}(f))^2 $. So,

$$I(t) = \mathcal{F}^{-1}((\mathcal{F}(f))^2)$$

Now $\mathcal{F}(f) = \pi e^{-|x|}$, so

$$(\mathcal{F}(f))^2 = \pi^2 e^{-|2x|} = 2\pi \cdot \pi/2e^{-|2x|} = \mathcal{F}\left(2\pi \frac{1}{1+(2t)^2}\right)$$

And so by the property of uniqueness for Fourier transform, we have that: $$I = 2\pi \frac{1}{1+4t^2}$$

Is this correct? And also, is there any other easier way to solve it? Also, I'm struggling oftenly to keep $x$ and $t$'s separated which makes me confused whether I should do the inverse fourier or the regular Fourier transform of a function. Is there any easy trick to distinguish these variables?

Answer 1

Sorry, too long for a comment

As @eyeballfrog provided the link in the comment, WolframAlpha gives $\displaystyle I(t)=\frac{2\pi}{4+t^2}$ (please, pay attention that not $\displaystyle \frac{2\pi}{1+4t^2}$).

I may suppose thay you define FT as $\displaystyle \hat f(k)=\int_{-\infty}^\infty f(x)e^{ikx}dx$. In this case you do get $\displaystyle\int_{-\infty}^\infty \frac{e^{ikx}}{1+x^2}dx=\pi e^{-|k|}$.

To find an answer, you could also use the Plancherel' theorem (here), which states (for the above chosen FT) that $$\int_{-\infty}^\infty f(x)g^*(x)dx=\frac{1}{2\pi}\int_{-\infty}^\infty \hat f(k)\hat g^*(k)dk$$ Taking $\,\displaystyle g(x)=\frac{1}{1+x^2},\,\, f(x)=\frac{1}{1+(x-t)^2}$, you get $\,\displaystyle \hat g(k)=\pi e^{-|k|}; \hat f(k)=\pi e^{-|k|}e^{ikt}$ $$I=\frac{\pi^2}{2\pi}\int_{-\infty}^\infty e^{-2|k|}e^{ikx}dk=\pi\, \Re\,\int_0^\infty e^{-2k}e^{ikx}dk=\pi\, \Re\,\frac{1}{2-it}=\frac{2\pi}{4+t^2}$$

Answer 2

As suggested by @Thomas Andrews. Do the partial fractions as: $$\frac{1}{(1+x^2)(1+(t-x)^2)}= \frac{Ax+B}{1+x^2}+\frac{Cx+D}{1+(t-x)^2}.....(1)$$ One gets $$A=\frac{2}{t(4+t^2)}, B=\frac{1}{4+t^2}, C=\frac{-2}{t(4+t^2)}, D=\frac{3}{(4+t^2)}.$$ While integrating (1) from $x=-\infty$ to $x=\infty$ the integral multiplied to $A$ vanishes and we get $$\implies I=\frac{1}{4+t^2} \left(\int_{-\infty}^{\infty} \frac{dx}{1+x^2}dx+ \int_{-\infty}^{\infty}\frac{dx}{1+(t-x)^,}+\int_{-\infty}^{\infty}\frac{2(t-x)dx}{1+(t-x)^2}\right).$$ Let $t-x=u$, then $$I=\frac{1}{4+t^2} \left(\int_{-\infty}^{\infty} \frac{dx}{1+x^2}dx+ \int_{-\infty}^{\infty}\frac{dx}{1+u^2}+\int_{-\infty}^{\infty}\frac{2udu}{1+u^2}\right).$$ The first two integrals give $\pi+\pi$ and the third one vaishes. Finally $$I=\frac{2\pi}{4+t^2}.$$