# Computing a double integral $\int_{-\infty}^\infty\int_{-\infty}^\infty\frac{f(t)}{1+{(x+g(t))}^2}dt\ dx$

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.

• Changing the order of integration is always one of the first things one should look at with double integrals. – Daniel Fischer Feb 2 '15 at 20:39
• @DanielFischer Sorry, I forgot to mention that I had not learned Fubini for indefinite integrals yet. I'll try with limits, though. – Hippalectryon Feb 2 '15 at 20:40
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• Without changing the order of integration this will be difficult. – PhoemueX Feb 2 '15 at 20:44
• 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. – JimmyK4542 Feb 2 '15 at 20:46

$\int_{-\infty}^\infty\int_{-\infty}^\infty\dfrac{f(t)}{1+{(x+g(t))}^2}dt~dx$
$=\int_{-\infty}^\infty\int_{-\infty}^\infty\dfrac{f(t)}{1+{(x+g(t))}^2}dx~dt$
$=\int_{-\infty}^\infty\left[f(t)\tan^{-1}(x+g(t))\right]_{-\infty}^\infty~dt$
$=\pi\int_{-\infty}^\infty f(t)~dt$