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Assume that you have two integrals

$$\int_0^\infty {1 \over x^2 + y^2} dx$$ and $$\int_{-\infty}^0 {-1 \over x^2 + y^2} dx$$

and you want to show that

$$\int_0^\infty {1 \over x^2 + y^2} dx + \int_{-\infty}^0 {-1 \over x^2 + y^2} dx = \int_0^\infty {1 \over x^2 + y^2} dx - \int_{-\infty}^0 {1 \over x^2 + y^2} dx = 0.$$

You are allowed to assume that $(x, y) \neq 0$.

It's quite obvious that the two integrals have the same value, but one still has to show that they have a finite value, because $\infty - \infty$ is not defined.

Is there a way to prove this without explicitly calculating the values of the integrals?

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    Why the downvote? – 2017-02-06
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    What should $\mathbb{R}_{\pm}$ mean in this context? You want to integrate with respect to $x,y$? – 2017-02-06
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    $\Bbb R_+ = (0, \infty), \Bbb R_- = (-\infty, 0)$. – 2017-02-06
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    Are you assuming $y \neq 0$? – 2017-02-06
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    @Borol Could you please specify your variable of integration in the first two integrals. It does not make sense at all. – 2017-02-06
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    Thanks, I edited the question, sorry. – 2017-02-06
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    For $y \neq 0$, $$\int_0^\infty \frac{1}{x^2 + y^2} dx = \frac{\pi}{2|y|},$$ which is not infinite. Where are your infinities coming from? – 2017-02-06
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    Please read the question again. I know that I can calculate the value of the integrals, but I wondered if there was a way to show that they are finite without calculating their values. – 2017-02-06
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    I don't understand the downvotes on this question at all. What did I do wrong? – 2017-02-06
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    To show convergence, use $$\frac1{x^2+y^2}\leqslant\frac1{y^2}\mathbf 1_{|x|\leqslant1}+\frac1{x^2}\mathbf 1_{|x|>1}$$ which shows that $$\int_0^\infty\frac{dx}{x^2+y^2}\leqslant\int_0^1\frac{dx}{y^2}+\int_1^\infty\frac{dx}{x^2}=\frac1{y^2}+1$$ and the same bound holds for the integral on $(-\infty,0)$. – 2017-02-06

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For $y \ne 0$: $\int_{\Bbb R_+} {1 \over x^2 + y^2} dx=\int_{0}^1 {1 \over x^2 + y^2} dx+\int_{1}^{\infty} {1 \over x^2 + y^2} dx$

The integral $ \int_{0}^1 {1 \over x^2 + y^2} dx$ makes no problems.

We have ${1 \over x^2 + y^2} \le {1 \over x^2 }$ and $\int_{1}^{\infty} {1 \over x^2 } dx$ converges.

Hence $\int_{\Bbb R_+} {1 \over x^2 + y^2} dx$ converges.

For $y = 0$: $\int_{\Bbb R_+} {1 \over x^2 } dx$ is divergent.