# Integral Homework: Solving $\int_0^{\infty} \frac{\log (x+1)}{x(x+1)} dx$

• Mr Davis 97
Taking the reciprocal helps this.Another idea would have been to split the domain and use different seriesfor each but I wanted to avoid that.I had expanding log(1-x) in mind, though using log(x) like Ray Vickson did worked also.dI=\frac{\log(x+1)}{x(x+1)}dx\\=-\frac{\log(\frac{1}{x+1})}{x(x+1)}dx\\=-\frac{\log(1+\frac{1}{x+1}-\frac{x+1}{x+1})}{x(x+1)}dx\\=-\fracf

## Homework Statement

##\displaystyle \int_0^{\infty} \frac{\log (x+1)}{x(x+1)} dx##

## The Attempt at a Solution

I tried to convert the log to a series, but that got be nowhere, since the resulting integral was divergent. Any hints on how to approach this?

• lurflurf

## Homework Statement

##\displaystyle \int_0^{\infty} \frac{\log (x+1)}{x(x+1)} dx##

## The Attempt at a Solution

I tried to convert the log to a series, but that got be nowhere, since the resulting integral was divergent. Any hints on how to approach this?

The indefinite integral can be expressed in terms of the non-elementary function ##\text{dilog}(\cdot)##, defined as
$$\text{dilog}(w) = \int_1^w \frac{\ln t}{1-t} \, dt$$
Some progress stems from integration by parts, using
$$u = \log(x+1) \;\;\text{and} \;\; dv = \frac{dx}{x(x+1)}$$

• jim mcnamara and lurflurf
hint find for ##k+k^\ast>-2##
##\int_0^\infty \log(x+1)(x+1)^{-k-2} dx##
then use geometric series
$$I=\int_0^\infty \frac{log(x+1)}{x(x+1)}dx\\ =\int_0^\infty \frac{log(x+1)}{[(x+1)-1](x+1)}dx\\ =\int_0^\infty \frac{log(x+1)}{[1-1/(x+1)](x+1)^2}dx\\ =\sum_{k=0}^\infty \int_0^\infty \log(x+1)(x+1)^{-k-2} dx$$

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I tried to convert the log to a series, but that got be nowhere, since the resulting integral was divergent. Any hints on how to approach this?
Your way is better than mine, but it helps to change variables.
let u=x/(x+1)
##
\int_0^\infty \frac{\log(x+1)}{x(x+1)}dx=\int_0^1 -\log(1-u)\frac{du}{u}
##
now expand in series.
I guess the two ways are similar, my way has an easy integral that looks hard and an easy series expansion that looks easy, your way has an easy integral that looks easy and an easy series expansion that looks hard. That was a fun integral.

• Mr Davis 97
The indefinite integral can be expressed in terms of the non-elementary function ##\text{dilog}(\cdot)##, defined as
$$\text{dilog}(w) = \int_1^w \frac{\ln t}{1-t} \, dt$$
Some progress stems from integration by parts, using
$$u = \log(x+1) \;\;\text{and} \;\; dv = \frac{dx}{x(x+1)}$$
https://en.wikipedia.org/wiki/Polylogarithm
That is nice. The change of variable t=1/(x+1) puts the given integral in the desired form. If we know in your notation (notation varies) ##\mathrm{dilog}(0)## we are done, otherwise we can use dilog identities or expand the logarithm in series as suggested. I was confused about the integration by parts at first I tried something similar and got stuck. Now I see the change of variable and integration by parts are equivalent. That is pretty weird.

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https://en.wikipedia.org/wiki/Polylogarithm
That is nice. The change of variable t=1/(x+1) puts the given integral in the desired form. If we know in your notation (notation varies) ##\mathrm{dilog}(0)## we are done, otherwise we can use dilog identities or expand the logarithm in series as suggested. I was confused about the integration by parts at first I tried something similar and got stuck. Now I see the change of variable and integration by parts are equivalent. That is pretty weird.

There is a wealth of material about the dilog (and other polylog functions), including values at some special arguments and the like. That allows the proposed definite integral to be performed in terms of known functions and their values, but whether the OP would be permitted to just go ahead and use those tabulated results is, or course, another matter entirely.

Your way is better than mine, but it helps to change variables.
let u=x/(x+1)
##
\int_0^\infty \frac{\log(x+1)}{x(x+1)}dx=\int_0^1 -\log(1-u)\frac{du}{u}
##
now expand in series.
I guess the two ways are similar, my way has an easy integral that looks hard and an easy series expansion that looks easy, your way has an easy integral that looks easy and an easy series expansion that looks hard. That was a fun integral.
This way allowed me to solve it. Where did you come up with the substitution ##\displaystyle u=\frac{x}{x+1}## though?

^As you point out we need to do something about the convergence.
Taking the reciprocal helps this.
Another idea would have been to split the domain and use different series
for each but I wanted to avoid that.
I had expanding log(1-x) in mind, though using log(x) like Ray Vickson did worked also.
##dI=\frac{\log(x+1)}{x(x+1)}dx\\
=-\frac{\log(\frac{1}{x+1})}{x(x+1)}dx\\
=-\frac{\log(1+\frac{1}{x+1}-\frac{x+1}{x+1})}{x(x+1)}dx\\
=-\frac{\log(1-\frac{x}{x+1})}{\frac{x}{x+1}(x+1)^2}(x+1)^2d\left(\frac{x}{x+1}\right)##

You could also use the substitution ##x = e^u-1## and then expand the integrand in powers of ##e^{-u}##.