4

Evaluate $$\int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x y) \ln (x y)} d x d y$$

I couldn't get very far on this one, so I would appreciate some help =)

My attempt so far (transcribed from the comments):

By extending it to the Dirichlet Eta Function I evaluated this integral to be $\ln 2$. I arrived at the identity below after differentiating a unit square integral expression for $\eta(2)$. $$\eta(s)\Gamma(s)=\int_0^1\int_0^1 \frac{(-\ln(xy))^{s-2}}{1+xy}dxdy$$ But I can't for the love of it solve it any differently than that. I would love to find an "elementary approach" if possible.

Lee David Chung Lin
  • 7,043
Flammable Maths
  • 538
  • 2
    From where did you get this integral? – Zacky Apr 25 '19 at 13:13
  • 1
    I constructed it myself. By extending it to the Dirichlet Eta Function I evaluated this integral to be ln(2). But I can't for the love of it solve it any differently than that. I would love to find an "elementary approach" if possible. – Flammable Maths Apr 25 '19 at 13:15
  • 5
    I see, that is nice. I think the answer is rather $-\ln 2$, also maybe it would be better if you include what you wrote above in the question, not everyone reads the comments. – Zacky Apr 25 '19 at 13:26
  • 1
    So put in other words, you want a different proof than using the following relation: $$\eta(s)\Gamma(s)=\int_0^1\int_0^1 \frac{(-\ln(xy))^{s-2}}{1+xy}dxdy$$ Which of course would be trivial putting in $s=1$ leading $-\ln 2$ to be the answer. I took that from here: https://en.wikipedia.org/wiki/Dirichlet_eta_function#Integral_representations – Zacky Apr 25 '19 at 13:34
  • Ayy, exactly. That is the identity I also arrived after differentiating a unit square integral expression for eta(2). – Flammable Maths Apr 25 '19 at 14:28
  • Have a look at https://arxiv.org/abs/math/0211148# – James Arathoon Apr 25 '19 at 14:32

2 Answers2

6

I think it is possible to avoid the Dirichlet eta function. Put\begin{equation*} I=\int_{0}^{1}\int_{0}^{1}\dfrac{1}{(1+xy)\ln(xy)}\,dxdy. \end{equation*} By symmetry \begin{equation*} I = 2\int_{0}^{1}\left(\int_{0}^{x}\dfrac{1}{(1+xy)\ln(xy)}\,dy\right)\, dx. \end{equation*} Via the substitution $y=\dfrac{z}{x}$ we get \begin{equation*} I = \int_{0}^{1}\left(\int_{0}^{x^2}\dfrac{1}{x(1+z)\ln(z)}\,dz\right)\,dx. \end{equation*} We have a double integral integrated over the domain $0<z<x^2, \, 0<x<1$. If we change the order of integration then we first have to integrate with respect to $x$ over $\sqrt{z}<x<1$ and then with respect to $z$ over $0<z<1.$ Thus \begin{gather*} I = 2\int_{0}^{1}\left(\int_{\sqrt{z}}^{1}\dfrac{1}{x(1+z)\ln(z)}\, dx\right)\, dz = 2\int_{0}^{1}\dfrac{1}{(1+z)\ln(z)}\left[\ln(x)\right]_{\sqrt{z}}^{1}\, dz =\\[2ex] \int_{0}^{1}\dfrac{-1}{1+z}\, dz = -\ln(2). \end{gather*}

JanG
  • 4,577
  • 2
    Wow, this bound interchanging is awesome! Maybe more directly:$$\int_{0}^{1} \int_{0}^{1} \frac{1}{(1+xy) \ln (xy)} dx dy\overset{xy=t}=\int_0^1 \int_0^y \frac{1}{y(1+t)\ln t} dt dy$$ $$=\int_0^1 \int_{t}^1 \frac{1}{y(1+t)\ln t}dy dt=\int_0^1 \frac{-\ln t}{(1+t)\ln t}dt=-\int_0^1 \frac{1}{1+t}dt=-\ln 2$$ – Zacky Apr 25 '19 at 19:17
  • $@$Zacky. That is really an improvement – JanG Apr 25 '19 at 19:38
  • Great one! Thank you :) – Flammable Maths Apr 25 '19 at 19:50
4

$$I(a)=\int_0^1\int_0^1 \frac{(xy)^{a}}{(1+xy)\ln(xy)}dxdy\Rightarrow I'(a)=\int_0^1\int_0^1\frac{(xy)^a}{1+xy}dxdy$$ $$\overset{xy=t}=\int_0^1\frac1y\int_0^y\frac{t^a}{1+t}dtdy=\sum_{n=0}^\infty (-1)^n \int_0^1 \frac1{y}\int_0^y t^{a+n}dtdy$$ $$\require{cancel}=\sum_{n=0}^\infty (-1)^n \int_0^1 \cancel{\frac1{y}}\frac{y^{a+n+\cancel1}}{a+n+1}dy=\sum_{n=0}^\infty (-1)^n \frac{1}{(a+n+1)^2}=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(a+n)^2}$$ We are looking to find $I(0)$, but we also have that $\lim\limits_{a\to \infty}I(a)=0$ (see also the comments bellow why not $a\to -\infty$), thus by Newton-Leibniz formula: $$I(0)=-(I(\infty)-I(0))=-\int^{\infty}_0 I'(a)da=-\sum_{n=1}^\infty(-1)^{n-1} \int^{\infty}_0 \frac{1}{(a+n)^2}da$$ $$=-\sum_{n=1}^\infty(-1)^{n-1}\left(-\frac{1}{a+n}\right)\bigg|^{\infty}_0 =-\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n}=-\ln 2 $$

Zacky
  • 27,674
  • 1
    (+1) Interesting solution! – mrtaurho Apr 26 '19 at 09:05
  • It's not $-\infty$ but $+\infty$. $xy<1$. Limits are not so easy to prove. – FDP Apr 28 '19 at 08:56
  • You mean I should've picked $I(\infty)=0$? Also since $x,y \in (0,1)$ isn't $xy>1$? – Zacky Apr 28 '19 at 09:04
  • When you multiply a positive number by something <1 the result is less than the previous number. Answer to your question: yes. – FDP Apr 28 '19 at 09:14
  • Prove that $\lim_{x\rightarrow +\infty}I(x)=0$ is not that easy and one needs to prove the other series limit is $0$ (probably more easy to do). – FDP Apr 28 '19 at 09:18
  • I got it now.. Do you think there's a way to comeback with this answer? If I choose $+\infty$ I get the answer to be $\ln 2$ which is not nice, or I have another mistake. – Zacky Apr 28 '19 at 09:39
  • The problem is to prove that the two limits are 0. Think about to prove that $\lim_{x\rightarrow +\infty}I(x)=0$. $\lim_{x\rightarrow +\infty} \sum_{k=0}^{\infty} u_n(x)$ is not always $\sum_{k=0}^{\infty} u_n(+\infty)$ if $u_n(\infty)=0$ – FDP Apr 28 '19 at 11:15
  • 1
    @Zacky: Your answer is nice don't delete it please. – FDP Apr 28 '19 at 18:01