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Consider the parameter integral

$$I(a)=\int_0^1\frac{\log(a+t^2)}{1+t^2}\,{\rm d}t\tag1$$

Where $a\in\mathbb{C}$. I am struggling to evaluate this integral in a closed-form.

However, first of all lets just concentrate on some particular values of $a$ for which I was actually able to evaluate the integral exactly

$$\begin{align} &a=0:&&\int_0^1\frac{\log(t^2)}{1+t^2}\,{\rm d}t=-2G\\ &a=1:&&\int_0^1\frac{\log(1+t^2)}{1+t^2}\,{\rm d}t=\frac{\pi}2\log(2)-G \end{align}$$

Here $G$ denotes Catalan's Constant. The first case is just one of many integral definitions of Catalan's Constant whereas the second case can be reduced to integrals of this type by the substitution $t=\tan(y)$. Furthermore WolframAlpha is capable of providing a closed-form for the case $a=-1$

$$a=-1:\int_0^1\frac{\log(t^2-1)}{t^2+1}\,{\rm d}t=\frac{\pi}4\log(2)+\frac{i\pi^2}4-G$$

It seems like the general anti-derivative of the case $a=-1$ can be expressed in terms of the Polylogarithm (the term can be found within the given link but is far to complicated to be included here).

For other values of $a$ I was not able to get anything done. I tried to expand the $\log$ and respectively the denominator as a series which ended up in an infinite summation of Hypergeometric Functions $($of the kind $_2F_1(1,k+1;k+2;-1/3)$ paired with a denominator depending on $k$$)$ I was not able to express explicit. Furthermore I tried to apply Feynman's Trick, i.e. differentiate w.r.t. to $a$ in order to get rid of the $\log$. The so occurring integral was easily evaluated by using partial fraction decomposition. Anyway I did not managed to find suitable borders for the integration w.r.t. $a$ afterwards. Applying a trigonometric substitution $($to be precise $t=\tan(x)$$)$ lead to the logarithmic term $\log(1+\cos^2(x))$ which I was not sure how to handle without invoking several powers of the cosine function $($i.e. by using the Taylor series expansion of the natural logarithm$)$.
The first approach aswell as the last one resulted in an infinite double summation. My knowledge about double sums, especially their evaluation, is quite weak. Maybe someone else is able to finish this up.


I have doubts that it is possible to derive an explicit closed-form expression for $I(a)$. Nevertheless for the case that the upper bound is given by $\infty$ instead of $1$ there actually exists a closed-form expression which makes me curious

$$I(a,b,c,g)=\int^\infty_0 \frac{\log(a^2+b^2x^2)}{c^2+g^2x^2}\,{\rm d}x = \frac{\pi}{cg}\log\left(\frac{ag+bc}{g}\right)\tag2$$

I am not familiar with the way this elegant relation was deduced since I just stumbled upon this one within this post.


I would highly appreciate an explicit expression for $I(a)$, maybe similar to the one given for $(2)$, even though I am not sure whether such a term exists. However, I am especially interested in the case $a=3$ for another integral I am working on right now.

Thanks in advance!

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    $\begingroup$ You can find an expression in terms of polylogarithms (which you might be able to simplify further). Possible approach: consider $I'(a)$, solve the integral and then try to integrate up. For $a=3$ you will get something like $I(3) = -G-\frac{1}{2} i \text{Li}_2\left(i-\frac{2}{-i+\sqrt{3}}\right)+\frac{1}{2} i \text{Li}_2\left(-i+\frac{2}{-i+\sqrt{3}}\right)+\frac{1}{4} \pi \log (2)-\frac{1}{3} i \pi \tan ^{-1}\left(e^{\frac{i \pi }{3}}\right)$ $\endgroup$
    – Winther
    Nov 11, 2018 at 19:18
  • $\begingroup$ @Winther I am a little bit confused about the borders of intgeration. To choose the upper bound as $a=3$ is clear to me but what about the other border? Should I not choose a value for which the expression vanishs in order to get a general anti-derivative, or am I mistake? Did you integrated from $0$ to $3$ back up? $\endgroup$
    – mrtaurho
    Nov 11, 2018 at 19:22
  • $\begingroup$ You know $I(0)$ so if you know any anti-derivative $J(a)$ of $I'(a)$ and then you have $I(a) = -2G + (J(a) - J(0))$ $\endgroup$
    – Winther
    Nov 11, 2018 at 19:24
  • $\begingroup$ @Winther Ah, okay. Thank you for pointing this out. $\endgroup$
    – mrtaurho
    Nov 11, 2018 at 19:25
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    $\begingroup$ With Mathematica I get for $a\geq 1$:$\frac{1}{16} i \left(-8 \text{Li}_2\left(-1-\frac{2}{\sqrt{a}-1}\right)+8 \text{Li}_2\left(-i \left(1+\frac{2}{\sqrt{a}-1}\right)\right)+8 \text{Li}_2\left(\frac{2 i}{\sqrt{a}+1}-i\right)-8 \text{Li}_2\left(\frac{2}{\sqrt{a}+1}-1\right)+4 i \pi \left(\log \left(\frac{2 i}{\sqrt{a}-1}+(1+i)\right)+\log \left(\frac{1+i}{a+(1+i) \sqrt{a}+i}\right)\right)+16 i C+\pi ^2-4 i \pi \log (2)\right)$ $\endgroup$ Nov 12, 2018 at 10:47

4 Answers 4

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$$\mathcal J(a,t)=\int_0^1 \frac{\ln(a+t(1+x^2))}{1+x^2}\mathrm dx\Rightarrow I(a)=\int_0^1\frac{\ln(a+x^2)}{1+x^2}\mathrm dx=\mathcal J(a-1,1)$$ $$ \frac{\mathrm d}{\mathrm dt}\mathcal J(a,t)=\int_0^1 \frac{\mathrm dx}{a+t+tx^2}=\frac{1}{\sqrt{t(t+a)}}\arctan\left(\sqrt{\frac{t}{t+a}}\right)$$ $$\mathcal J(a,0)=\frac{\pi\ln a}{4}\Rightarrow \mathcal J(a,1)=\underbrace{\int_0^1 \frac{1}{\sqrt{t(t+a)}}\arctan\left(\sqrt{\frac{t}{t+a}}\right)\mathrm dt}_{=J}+\frac{\pi\ln a}{4}$$ Now via the substitution $\displaystyle{\sqrt{\frac{t}{t+a}}=x\Rightarrow \frac{\mathrm dt}{\sqrt{t(t+a)}}=\frac{2}{1-x^2}dx}$ we get: $$J=2\int_0^\frac{1}{\sqrt{1+a}}\frac{\arctan x}{1-x^2}\mathrm dx \overset{x=\frac{1-y}{1+y}}=\int_{\frac{\sqrt{1+a}-1}{\sqrt{1+a}+1}}^1\frac{\arctan\left(\frac{1-y}{1+y}\right)}{y}\mathrm dy$$ $$=\frac{\pi}{4}\int_{\frac{\sqrt{1+a}-1}{\sqrt{1+a}+1}}^1\frac{\mathrm dy}{y}-\int_0^1 \frac{\arctan y}{y}\mathrm dy+\int^{\frac{\sqrt{1+a}-1}{\sqrt{1+a}+1}}_0\frac{\arctan y}{y}\mathrm dy$$ $$\Rightarrow \mathcal J(a,1)=\frac{\pi}{4} \ln\left(\frac{\sqrt{a+1}+1}{\sqrt{a+1}-1}\right)-\mathrm G+\operatorname{Ti}_2\left(\frac{\sqrt{a+1}-1}{\sqrt{a+1}+1}\right)+\frac{\pi}{4}\ln a$$ $$\Rightarrow \boxed{I(a)=\int_0^1 \frac{\ln(a+x^2)}{1+x^2}dx=\frac{\pi}{2}\ln(\sqrt a+1)-\mathrm G+\operatorname{Ti}_2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)}$$ Where $\operatorname{Ti}_2(x)$ is the inverse tangent integral and $\mathrm G$ is Catalan's constant.


Some nice results that follows: $$\boxed{I(3)=\int_0^1\frac{\ln(3+x^2)}{1+x^2}\mathrm dx=\frac{\pi}{4}\ln 2+\frac{\pi}{6}\ln(2+\sqrt 3)-\frac13\mathrm G}$$ $$\boxed{I\left(\frac13\right)=\int_0^1 \frac{\ln\left(\frac13 +x^2\right)}{1+x^2}\mathrm dx=\frac{\pi}4 \ln \left(\frac23\right)+\frac{\pi}{3}\ln(2+\sqrt 3)-\frac53\mathrm G}$$

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    $\begingroup$ Impressive! Thank you kindly for this! $\endgroup$
    – mrtaurho
    Apr 20, 2019 at 19:38
  • $\begingroup$ @mrtaurho I had some free time and made the approach more clean, hope it's better now. $\endgroup$
    – Zacky
    Jul 24, 2019 at 14:26
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    $\begingroup$ As I already upvoted and accepted your answer there is not really much left which I could do to show my appreciation. Thank you! $\endgroup$
    – mrtaurho
    Jul 24, 2019 at 14:42
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I get $$ I'(a) = \int_0^1 \dfrac{dt}{(t^2+a)(t^2+1)} = {\frac {\pi\,\sqrt {a}-4\,\arctan \left( {\frac {1}{\sqrt {a}}} \right) }{ 4 \left( a-1 \right) \sqrt {a}}} $$ Integrating this using Maple produces a rather complicated expression, which seems to work for $0 < a < 1$ (a different branch should be used after $a=1$): $$\frac{i}{2}{\it dilog} \left( {\frac {-2\,\sqrt {a}+1-i+ \left( 1+i \right) a }{a+1}} \right) -\frac{i}{2}{\it dilog} \left( {\frac {2\,\sqrt {a}+1+i+ \left( 1-i \right) a}{a+1}} \right) +\frac{\pi}{4}\,\ln \left( 1-\sqrt {a } \right) +\frac{\pi}{4}\,\ln \left( 1+\sqrt {a} \right) -\frac{1}{2}\,\arctan \left( {\frac {1}{\sqrt {a}}} \right) \ln \left( -\sqrt {2} \left( 1 +\sqrt {a} \right) \sqrt {a+1}+2\,a+2 \right) +\frac{1}{2}\,\arctan \left( { \frac {1}{\sqrt {a}}} \right) \ln \left( \sqrt {2} \left( -1+\sqrt {a } \right) \sqrt {a+1}+2\,a+2 \right) -\frac{1}{2}\,\arctan \left( {\frac {1}{ \sqrt {a}}} \right) \ln \left( \sqrt {2} \left( 1+\sqrt {a} \right) \sqrt {a+1}+2\,a+2 \right) +\frac{1}{2}\,\arctan \left( {\frac {1}{\sqrt {a}}} \right) \ln \left( -\sqrt {2} \left( -1+\sqrt {a} \right) \sqrt {a+1 }+2\,a+2 \right) +2\,i\arctan \left( {\frac {1}{\sqrt {a}}} \right) \arctan \left( {\frac {-1+\sqrt {a}}{-\sqrt {2}\sqrt {a+1}+\sqrt {a}+1 }} \right) +2\,i\arctan \left( {\frac {1}{\sqrt {a}}} \right) \arctan \left( {\frac {1+\sqrt {a}}{\sqrt {2}\sqrt {a+1}+\sqrt {a}-1}} \right) -i\arctan \left( {\frac {1}{\sqrt {a}}} \right) \pi-\frac{i}{4}{\pi}^ {2}-{\it Catalan} $$

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  • $\begingroup$ I guess dilog refers to the Dilogarithm $\operatorname{Li}_2$? Howsoever do you see any way how to simplify this expression a little bit further? Moreover this term is from a my point of view not usable since I am looking especially for the case $a=3$ even though it is astonishing that at least Maple was able to provide a closed-form. What exactly do you mean with a different branch should be used after $a=1$? I am aware of the rather small radii of convergence of the involved functions - at least within their series representation - but to what branch do you exactly refer? $\endgroup$
    – mrtaurho
    Nov 11, 2018 at 20:26
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    $\begingroup$ "Branch" in the sense of complex analysis. Although the singularity of $I'(a)$ at $a=1$ is removable, the closed form for its antiderivative has terms $-\arctan \left( {\frac {1}{\sqrt {a}}} \right) \ln \left( \sqrt {2} \sqrt {1+a}-1-i- \left( 1-i \right) \sqrt {a} \right) $ and $\frac{\pi}{4}\,\ln \left( -\sqrt {a}+1 \right) $ that have a branch point at $a=1$. Care must be used in selection of the branches of these multivalued functions. $\endgroup$ Nov 11, 2018 at 21:27
  • $\begingroup$ Ah okay. That makes sense. To be honest I have not took a closer look on the formula you provided before I wrote my comment. Exspecially the critical points $a=0$ and $a=1$ are clearly visible within the inverse tangent aswell as within the logarithm which makes my confusion in the first place kind of redundant. While examining the different components of the anti-derivative I noticed that there are pairwise quite simmetric but unfortunately I do not know how to use this fact to simplify the terms... $\endgroup$
    – mrtaurho
    Nov 11, 2018 at 21:32
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    $\begingroup$ Maple's dilog is not quite the same as $\text{Li}_2$: $\text{dilog}(z) = \text{Li}_2(1-z)$. $\endgroup$ Nov 12, 2018 at 1:15
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    $\begingroup$ In Maple you can convert to polylogs by command convert: convert(expression,polylog) $\endgroup$ Nov 12, 2018 at 10:46
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Let $a>-1$ be a real number. Then

$$\int_0^1 \frac{\log(1+a^2x^2)}{1+x^2}\textrm{d}x=\frac{\pi}{2}\log(1+a)-G+\text{Ti}_2\left(\frac{1-a}{1+a}\right),$$

where $G$ is the Catalan's constant and $\displaystyle \text{Ti}_2(x)=\int_0^x \frac{\arctan(t)}{t}\textrm{d}t$ is the inverse tangent integral.

Thanks to Cornel for this way of writing the closed-form of the integral.

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    $\begingroup$ Interesting! (+1) Do you have a proof too? $\endgroup$
    – mrtaurho
    Jul 23, 2019 at 21:05
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    $\begingroup$ @mrtaurho Cornel says all we have to do is to differentiate with respect to $a$, then integrate with respect to $x$, then with respect to $a$ and play with simple integration by parts to get this form. Most probably if you do that you'll get this form without any difficulty. $\endgroup$ Jul 23, 2019 at 21:09
  • $\begingroup$ As this is precisely what Zacky did in his answer, it seems like the integral in the end breaks down easier than both of us expected. $\endgroup$
    – mrtaurho
    Jul 23, 2019 at 21:14
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    $\begingroup$ It's sad to see these days results spitted out without a proof, I mean instead of writing that comment from above you could've just put that in practice into the answer, it takes less than $5$ minutes. $\endgroup$
    – Zacky
    Jul 23, 2019 at 21:18
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    $\begingroup$ @mrtaurho you can just let $\frac{1-x}{1+x}=t$ in the representation of $J$ from my answer, then use $\arctan\left(\frac{1-x}{1+x}\right)=\frac{\pi}{4}-\arctan x$ to get this closed form (or a similar one) if u want. $\endgroup$
    – Zacky
    Jul 23, 2019 at 21:19
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$$\int_0^1\frac{\ln(a^2+x^2)}{1+x^2}dx=\int_0^\infty\frac{\ln(a^2+x^2)}{1+x^2}dx-\underbrace{\int_1^\infty\frac{\ln(a^2+x^2)}{1+x^2}dx}_{x\to 1/x}$$

$$=\int_0^\infty\frac{\ln(a^2+x^2)}{1+x^2}dx-\int_0 ^1\frac{\ln(\frac{1+a^2x^2}{x^2})}{1+x^2}dx$$

$$=\int_0^\infty\frac{\ln(a^2+x^2)}{1+x^2}dx-\int_0 ^1\frac{\ln(1+a^2x^2)}{1+x^2}dx-2\int_0^1\frac{-\ln(x)}{1+x^2}dx$$ $$=\pi\ln(1+a)-\int_0 ^1\frac{\ln(1+a^2x^2)}{1+x^2}dx-2G$$ $$=\pi\ln(1+a)-I-2G,$$

where the first integral was evaluated by differentiation under the integral sign and the third integral is the integral representation of the Catalan constant. To find $I$, let $$I(b)=\int_0 ^1\frac{\ln(1+b^2x^2)}{1+x^2}dx$$

and notice that $I(a)=I$ and $I(0)=0,$

$$I'(b)=\int_0^1\frac{2bx^2}{(1+x^2)(1+b^2x^2)}dx=\frac{4\arctan(b)-b\pi}{2(1-b^2)}.$$

Now integrate both sides from $b=0$ to $b=a,$

$$\int_0^a I'(b)db=I(b)|_0^a=I(a)-I(0)=I-0=\int_0^a \frac{4\arctan(b)-b\pi}{2(1-b^2)}db$$

$$\overset{b=(1-x)/(1+x)}{=}\frac{\pi}{2}\int_{\frac{1-a}{1+a}}^1\frac{dx}{1+x}dx-\int_{\frac{1-a}{1+a}}^1\frac{\arctan(x)}{x}dx$$

$$I=\frac{\pi}{2}\ln(1+a)-\left(\int_0^1-\int_0^\frac{1-a}{1+a}\right)\frac{\arctan(x)}{x}dx$$

$$I=\frac{\pi}{2}\ln(1+a)-G+\text{Ti}\left(\frac{1-a}{1+a}\right),$$

and so

$$\int_0^1\frac{\ln(a^2+x^2)}{1+x^2}dx=\frac{\pi}{2}\ln(1+a)-G-\text{Ti}\left(\frac{1-a}{1+a}\right).$$

Note that $\text{Ti}(-x)=-\text{Ti}(x)$ since $\arctan(-x)=-\arctan(x).$

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