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I am struggling to understand when an integral and an infinite sum can be switched around. I want to know if they can be switched in the following case:

$$\sum_{a=1}^{\infty} \int_{0}^{\infty} \frac{\cos (t x)}{x^{2}+a^{2}}\mathop{dx}$$

I don’t need the reader to evaluate the sum/integral.

phi-rate
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  • Does this help? Also look at this.Furthermore. – Bumblebee Jan 06 '22 at 04:30
  • Are you sure the lower index started at $a=0$? The improper integral $\int_0^{\infty}\frac{\cos(tx)\mathrm{d}x}{x^2}$ diverges. – Matthew H. Jan 06 '22 at 04:33
  • Thank you so much for pointing that out! I shall edit it now @MatthewH. – phi-rate Jan 06 '22 at 04:35
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    If $t=0$, you get a divergent sum. If $t\neq 0$, then using a change of variables and integration by parts $$ \int_0^{ + \infty } {\frac{{\cos (tx)}}{{x^2 + a^2 }}dx} = \frac{1}{a}\int_0^{ + \infty } {\frac{{\cos (taz)}}{{z^2 + 1}}dz} = \frac{1}{{a^2 }}\frac{2}{t}\int_0^{ + \infty } {\frac{{z\sin (taz)}}{{(z^2 + 1)^2 }}dz} . $$ Now $$\left| {\frac{1}{{a^2 }}\frac{2}{t}\int_0^{ + \infty } {\frac{{z\sin (taz)}}{{(z^2 + 1)^2 }}dz} } \right| \le \frac{1}{{a^2 }}\frac{2}{{\left| t \right|}}\int_0^{ + \infty } {\frac{z}{{(z^2 + 1)^2 }}dz = } \frac{1}{{\left| t \right|}}\frac{1}{{a^2 }}. $$ – Gary Jan 06 '22 at 04:52

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