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For fun, I was looking at the following Putnam-Style problem the other day on this page: (It is problem B2)

Evaluate the integral $$I(a,b)=\int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{x\left(e^{ax}+1\right)\left(e^{bx}+1\right)}dx$$ for $a>b>0$.

Since it is suppose to be a putnam style problem, there certainly must be a large number of not too difficult approaches. I have a solution, but I really don't think it is an intended one since it uses some facts about the Dirichlet Eta Function and a Mellin transform which relates it to an integral and the Gamma Function. (I posted it below)

I am asking this question because I would like to see different solutions to this problem. Specifically I would like to know how to solve it using "differentiation under the integration sign," since I am trying to learn that technique. (An explanation for why we can bring the derivative inside the integral would also be very useful!)

Thanks a lot,

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Two simple remarks: because of the $\frac{dx}{x}$, we can see that $I(\lambda a, \lambda b) = I(a,b)$ (by a change of variable $u = \lambda x$), so $I(a,b)$ only depends on $\frac{a}{b}$. And we obviously have $I(1,1) = 0$. This does not quite get you the solution, but I could not resist the urge to make that comment. – Joel Cohen May 31 '11 at 16:09
I request you to accept an answer. – user9413 Jun 7 '11 at 16:17
@Chandru: I was contemplating putting a bounty on this question to get more answers. But I accepted Didiers answer. (I couldn't accept Sivaram's because I don't know the justification for switching. I think there should be an easy reason...) – Eric Naslund Jun 7 '11 at 16:37
Dear, Eric no problem. Please don't accept an answer just because I said it. I felt that you had forgotten so i thought of reminding you. Anyway you have a valid reason for not accepting an answer. – user9413 Jun 7 '11 at 16:42
@Chandru: You did remind me however of a problem where I forgot to accept a good answer. (A PDE problem) I also decided to accept the top voted answers for the FAQ style questions I asked, but I am not sure if that was necessary. – Eric Naslund Jun 7 '11 at 16:44
up vote 12 down vote accepted

Here is a first solution, which uses no complex variable nor any differentiation, just plain real analysis. A second solution follows, which uses Fubini theorem.

For every positive $t$, let $I_t(a,b)$ denote the integral of the same function than in the integral which defines $I(a,b)$, but integrated from $t$ to $+\infty$. Then $$ I_t(a,b)=\int_{t}^{+\infty}\frac{\mathrm{d}x}{x\left(\mathrm{e}^{bx}+1\right)}-\int_{t}^{+\infty}\frac{\mathrm{d}x}{x\left(\mathrm{e}^{ax}+1\right)}= \int_{bt}^{at}\frac{\mathrm{d}x}{x\left(\mathrm{e}^{x}+1\right)}, $$ where the first decomposition of $I_t(a,b)$ into two parts is legal because we got rid of the problem at $x=0$ and the second formula stems from the changes of variables $x\leftarrow bx$ and $x\leftarrow ax$ in the two parts.

Extracting $1/(2x)$ from the last fraction and integrating, one gets $$ I_t(a,b)=\frac12\log\left(\frac{a}b\right)- \int_{bt}^{at}\frac{\mathrm{e}^{x}-1}{2x\left(\mathrm{e}^{x}+1\right)}\mathrm{d}x=\frac12\log\left(\frac{a}b\right)+O(t), $$ where the $O(t)$ term when $t\to0$ comes from the fact that the function in the last integral is bounded around $x=0$ and from the fact that this integral is over an interval of length $at-bt=O(t)$. Finally, when $t\to0$, $I_t(a,b)\to I(a,b)$ hence $$ I(a,b)=\frac12\log\left(\frac{a}b\right). $$

Here is a second solution, based on Fubini theorem. Introduce $$ u(c,x)=\frac{\mathrm{e}^{cx}}{(1+\mathrm{e}^{cx})^2}=\frac{\partial}{\partial c}\left(\frac{-1}{x(1+\mathrm{e}^{cx})}\right). $$ One sees that $$ \frac{\mathrm{e}^{ax}-\mathrm{e}^{bx}}{x\left(\mathrm{e}^{ax}+1\right)\left(\mathrm{e}^{bx}+1\right)}=\frac1{x\left(\mathrm{e}^{bx}+1\right)}-\frac1{x\left(\mathrm{e}^{ax}+1\right)}=\int_b^au(c,x)\mathrm{d}c, $$ hence $$ I(a,b)=\int_{0}^{\infty}\left(\int_b^au(c,x)\mathrm{d}c\right)\mathrm{d}x. $$ Since $u\ge0$, Fubini-Tonelli theorem says one can interchange the order of integration, hence $$ I(a,b)=\int_b^aU(c)\mathrm{d}c,\qquad U(c)=\int_{0}^{\infty}u(c,x)\mathrm{d}x. $$ Obviously, $$ u(c,x)=\frac{\partial}{\partial x}\left(\frac{-1}{c(1+\mathrm{e}^{cx})}\right),\qquad U(c)=\left.\frac{-1}{c(1+\mathrm{e}^{cx})}\right|^{x=+\infty}_{x=0}=\frac1{2c}, $$ and the value of $I(a,b)$ follows.

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I like this solution a lot, thanks! It is very simple too. – Eric Naslund May 31 '11 at 22:41
@Eric: You are welcome. – Did May 31 '11 at 22:44
I accept this answer. – Eric Naslund Jun 7 '11 at 16:37
...And you get as a bonus a second solution, closer to Sivaram's approach than my first one and based on Fubini theorem for positive functions. (I was waiting for somebody else to propose this alternative proof, but, since the post is now going to disappear from the first page and nobody proposed anything, I do it myself.) – Did Jun 7 '11 at 19:22

$$I(a,b)=\int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{x\left(e^{ax}+1\right)\left(e^{bx}+1\right)}dx = \int_{0}^{\infty} \left(\frac{1} {x\left(e^{bx}+1\right)}- \frac{1} {x\left(e^{ax}+1\right)} \right)dx$$

$$\frac{\partial I}{\partial a} = \int_0^{\infty} \frac{e^{ax}}{(1+e^{ax})^2} dx$$

Let $t = e^{ax}$. We get

$$\frac{\partial I}{\partial a} = \frac1{a} \int_1^{\infty} \frac{dt}{(1+t)^2} = \frac{1}{2a}$$

Similarly, $$\frac{\partial I}{\partial b} = - \frac1{b} \int_1^{\infty} \frac{dt}{(1+t)^2} = -\frac{1}{2b}$$

Hence, $I(a,b) = \frac1{2} \left( \log(a) - \log(b) \right) + c$.

Further $I(a,a) = 0$ and hence $c = 0$. Hence,

$$I(a,b) = \frac1{2} \log \left(\frac{a}{b}\right)$$

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This is a very nice and simple solution, if we ignore a minor issue. The right hand side of the first line is invalid since both integrals diverge. It would be better to put them all in a single integral operator. – Sangchul Lee May 31 '11 at 15:40
@sos440: Thanks. I dint realize the individual integrals diverge. I have now edited it and I think now the solution is correct. – user17762 May 31 '11 at 15:44
How do we justify bringing the partial derivative inside the integral? – Eric Naslund May 31 '11 at 16:08

Here is my solution with the eta function:

Let $\text{Re}(s)>1$, and consider $$\int_{0}^{\infty}\frac{\left(e^{ax}-e^{bx}\right)x^{s-1}}{\left(e^{ax}+1\right)\left(e^{bx}+1\right)}dx=\int_{0}^{\infty}\frac{x^{s-1}}{e^{bx}+1}dx-\int_{0}^{\infty}\frac{x^{s-1}}{e^{ax}+1}.$$ Recall that for $\text{Re}(s)>1$ we have $$\Gamma(s)\eta(s)=\int_{0}^{\infty}\frac{t^{s-1}}{e^{t}+1}dt$$ where $\eta(s)$ is the dirichlet eta function. Hence, by a substitution, the above expression is $$\frac{\Gamma(s)\eta(s)}{b^{s}}-\frac{\Gamma(s)\eta(s)}{a^{s}}=\Gamma(s)\eta(s)\left(e^{-s\log b}-e^{-s\log a}\right)$$ $$=\Gamma(s)\eta(s)\left(s\left(\log a-\log b\right)+s^{2}\left(\cdots\right)+\cdots\right)$$ Since $\Gamma(s)$ has a simple pole with residue $1$ at $s=0$, and $\eta(0)=\left(1-2\right)\zeta(0)=\frac{1}{2}$, we have $$\lim_{s\rightarrow0}\int_{0}^{\infty}\frac{e^{ax}-e^{bx}}{\left(e^{ax}+1\right)\left(e^{bx}+1\right)}x^{s-1}dx=\frac{1}{2}\log\left(\frac{a}{b}\right).$$ Switching the limit and the integral is then justifiable using the dominated convergence theorem so that we conclude $$I(a,b)=\frac{1}{2}\log\left(\frac{a}{b}\right).$$

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Smokin' hot. XD – Patrick Da Silva Jun 24 '11 at 6:03

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