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Let $f,g$ be continuous, with $f$ integrable.

How can one evaluate $\displaystyle\int_{-\infty}^\infty\int_{-\infty}^\infty\dfrac{f(t)}{1+{(x+g(t))}^2}dt\ dx$ ?

Any hint would be welcome.

I have not learned Funibi/Tonelli for indefinite integrals yet.

PS : To those who are voting to close as "too broad", would you mind explaining what is too broad in my question ? I am just asking for a way to evaluate the integral, or a hint that would lead me to the solution.

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    $\begingroup$ Changing the order of integration is always one of the first things one should look at with double integrals. $\endgroup$ – Daniel Fischer Feb 2 '15 at 20:39
  • $\begingroup$ @DanielFischer Sorry, I forgot to mention that I had not learned Fubini for indefinite integrals yet. I'll try with limits, though. $\endgroup$ – Hippalectryon Feb 2 '15 at 20:40
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    $\begingroup$ I have changed the formatting of the title so as to make it take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – AlexR Feb 2 '15 at 20:41
  • $\begingroup$ Without changing the order of integration this will be difficult. $\endgroup$ – PhoemueX Feb 2 '15 at 20:44
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    $\begingroup$ Try the substitution $u = x+g(t)$, $v = t$. The determinant of the Jacobian of this transformation is $1$. Although this is more or less the same thing as changing the order of integration. $\endgroup$ – JimmyK4542 Feb 2 '15 at 20:46
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$\int_{-\infty}^\infty\int_{-\infty}^\infty\dfrac{f(t)}{1+{(x+g(t))}^2}dt~dx$

$=\int_{-\infty}^\infty\int_{-\infty}^\infty\dfrac{f(t)}{1+{(x+g(t))}^2}dx~dt$

$=\int_{-\infty}^\infty\left[f(t)\tan^{-1}(x+g(t))\right]_{-\infty}^\infty~dt$

$=\pi\int_{-\infty}^\infty f(t)~dt$

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