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I have integral $\int_0^\infty \frac{2a^2}{x^2 + a^2} \frac{\sin x}{x}dx$ after the next transformation $\frac{\sin x}{x} = \frac{1}{x} \int_0^1 x\cos(yx) dy = \int_0^1 \cos(yx) dy$ I get a double integral $\int_0^\infty \frac{2a^2}{x^2 + a^2} (\int_0^1 \cos(yx) dy)dx$.

Am I correct in assuming that I can present it in this form $\int_0^1 dy\int_0^\infty \frac{2a^2}{x^2 + a^2}\cos(yx) dx$ ? where the general function f(x,y) is integrated by x (with appropriate integration bounds) and only then integrated by y? As far as I know I correct, but I am still confused by the fact that the integral $ \int_0^1 \cos(yx) dy $ is inside of this one $\int_0^\infty \frac{2a^2}{x^2 + a^2} dx$ and thats why I can't integrate function $\cos(yx)$ by x until I integrate it by y over [0 1].

Varga
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1 Answers1

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$\int_0^\infty \dfrac{2a^2}{x^2 + a^2} \int_0^1 \cos(yx) dydx$ can be rearranged as $$\int_0^\infty \int_0^1\frac{2a^2}{x^2 + a^2} \cos(yx) dydx$$ Now we can change the order of integration as $$ \int_0^1\int_0^\infty\frac{2a^2}{x^2 + a^2} \cos(yx) dxdy$$ No further simplification is possible than $$ 2a^2\int_0^1\int_0^\infty\frac{\cos(yx)}{x^2 + a^2} dxdy$$

neo
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