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I would like to get the following integral:

$$\int -\frac{\log(a^2+x^2)}{(a^2+x^2)}dx \quad \text{or} \quad \int_{t}^{+\infty}-\frac{\log(a^2+x^2)}{(a^2+x^2)}dx$$

where $t>0$.

I used WolframAlpha to compute, and I got following expression:

enter image description here

I am just wondering why some complex value i appears? Any ideas of how to get the closed form of integral from t to +inf ? I guess I need to give more specification to wolframalpha to compute ?

Ѕᴀᴀᴅ
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Jiadong
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  • An integral from $t$ to $\infty$ is no better than an indefinite integral when it comes to solvability. An integral from $0$ to $\infty$ may be a bit better. – Parcly Taxel Sep 29 '22 at 06:17
  • my goal is to get the closed form of integration from t1 to t2 (real positive values). but I am just confused about the complex value i of the output... do you have any ideas which can guide me? thanks – Jiadong Sep 29 '22 at 06:20
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    @Jiadong: the three last terms are complex numbers when x,a are real do you know that the sum of complex numbers could be a real one? – FDP Sep 29 '22 at 06:40
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    Wolfram alpha is using the polylogarithmic integral for convenience. The complex numbers are just needed to make the polylogarithms applicable, but the whole expression will be entirely real. – FShrike Sep 29 '22 at 06:52
  • @FShrike, ya, I know complex numbers can add up to get real numbers, but I haven't check this case. Maybe I will plug in some numbers to test. – Jiadong Sep 29 '22 at 06:57
  • If WolfromAlpha is right (I believe), last three terms definitely will add up to get a real number for any real x. Ya this might be a convenient way to represent..., but polylog is not easy to evaluate, but I accept this form if no better form out there. – Jiadong Sep 29 '22 at 07:14
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    Just a stupid question : what would you do by yourself (without WA) beside introducing in particular $x^2+a^2=(x+ia)(x-ia)$ before partial fraction decomposition ? – Claude Leibovici Sep 29 '22 at 07:17
  • I meant to refer to what Claude is hinting. Wolfram Alpha is using complex numbers for convenience so that it can get an easier to integrate expression in terms of polylogs – FShrike Sep 29 '22 at 07:43

1 Answers1

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For real $x$ and $a$ it is possible to rewrite the result using only real-valued functions. After copious use of the function relations $$\arctan z=-\frac i2\log\frac{i-z}{i+z}$$ $$\operatorname{Li}_2(e^{it})=\frac{\pi^2}6-\frac{\pi t}2+\frac{t^2}4+i\operatorname{Cl}_2(t)$$ where $\operatorname{Cl}_2$ is the classical Clausen function, the simplified result is $$\int_0^x\frac{\log(t^2+a^2)}{t^2+a^2}\,dt=\frac1a\left(2\log2a\arctan\frac xa-\operatorname{Cl}_2\left(\pi-2\arctan\frac xa\right)\right)$$ The integral over $[0,\infty)$ is $\frac{\pi\log2a}a$; cf. here.

Parcly Taxel
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