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)...

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

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

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

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

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    $\begingroup$ See here. $\endgroup$ – Eff Jan 29 '15 at 7:35
  • $\begingroup$ @Eff Thanks :) I'm checking it out, right now $\endgroup$ – The Artist Jan 29 '15 at 7:36
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    $\begingroup$ 1. You are missing a factor $(-)^k$ in your series. 2. For general series/integral, it is illegal to exchange limit of integral and sum. 3. However, $\frac{\log x}{x^{4k-2}} - \frac{\log x}{x^{4k}}$ are non-negative over $[1,\infty)$. A long as you manipulate the terms of your series in units of pair, the exchange is legal in this particular case. $\endgroup$ – achille hui Jan 29 '15 at 7:43
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    $\begingroup$ In general, if you have a sequence of function $f_n(z)$ which are non-negative over some interval $(a,b)$ where $a,b$ can be finite or infinite, $$\int_a^b \sum_n f_n(z) dz = \sum_n \int_a^b f_n(z) dz$$ in the sense that if one side converges, so does the other side and they have the same limit. If one side diverges, so does the other side. Among all test of exchange of limit, this is the one most easiest to use. $\endgroup$ – achille hui Jan 29 '15 at 8:06
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    $\begingroup$ Let $t=\dfrac1x$ $\endgroup$ – Lucian Jan 29 '15 at 11:40

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