Can we interchange the Integral and Summation when a limit is $\infty$?

Question

I was trying to Evaluate the Integral:

$$\Large{I=\int_1^{\infty} \frac{\ln x}{x^2+1} dx}$$

$$\color{#66f}{{\frac{1}{x^2+1} = \frac{1}{x^2\left(1+\frac{1}{x^2}\right)}=\frac{1}{x^2}\cdot \frac{1}{1+x^{-2}}=\frac{1}{x^2} \sum_{n=0}^{\infty} \left(\frac{1}{-x^{2}}\right)^{n}}}$$

$${I=\int_1^{\infty} \left(\frac{\ln x}{x^2}-\frac{\ln x}{x^4}+\frac{\ln x}{x^6}+\cdots\right)dx}=-\int_1^{\infty} \sum_{k=1}^{\infty} \frac{\ln x}{(-1)^kx^{2k}} dx$$

Now I would like to interchange the integral and the summation, like :

$$-\int_1^{\infty} \sum_{k=1}^{\infty} \frac{\ln x}{(-1)^kx^{2k}} dx=-\sum_{k=1}^{\infty} \int_1^{\infty} \frac{\ln x}{(-1)^kx^{2k}} dx$$

But I'm not sure if I can do that when $\infty$ is present (not real)...

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$$\text{I know that:}$$

$$\bbox[8pt, border: solid 2pt crimson]{-\int_1^{b} \sum_{k=1}^{a} \frac{\ln x}{(-1)^kx^{2k}} dx=-\sum_{k=1}^{a} \int_1^{b} \frac{\ln x}{(-1)^kx^{2k}} dx}$$

$$\color{crimson}{\text{Where a, b is real}}$$

But is it the same when $a,b=\infty$?

Answer 1

$$\int\limits_{1}^{\infty}\frac{\log\left(x\right)}{x^2+1}\,dx= \int\limits_{0}^{\infty}\frac{ue^u}{e^{2u}+1}\,du$$ $$\int\limits_{0}^{\infty}\frac{ue^u}{e^{2u}+1}\,du=\sum_{n=1}^{\infty}\left(-1\right)^{n+1}\int\limits_{0}^{\infty}e^{-2nu+u}u\,du$$ $$\sum_{n=1}^{\infty}\left(-1\right)^{n+1}\int\limits_{0}^{\infty}e^{-2nu+u}u\,du=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{\left(2n-1\right)^2}\int\limits_{0}^{\infty}e^{-t}t\,dt=\sum_{n=1}^{\infty}\frac{\left(-1\right)^{n+1}}{\left(2n-1\right)^2}=\textbf{G}$$ Where $\textbf{G}$ is Catalan's constant. $$\boxed{\int\limits_{1}^{\infty}\frac{\log\left(x\right)}{x^2+1}\,dx=\textbf{G}}$$