Can you help me find the value of the integral
$\int_{1}^{2}\frac{dx}{1+x+\ln x}$
Thank you
Can you help me find the value of the integral
$\int_{1}^{2}\frac{dx}{1+x+\ln x}$
Thank you
According to Maple, there is no closed form either for the antiderivative or the definite integral. In this "pure transcendental" case, Maple's implementation of the Risch algorithm is complete, so there is no elementary antiderivative (this could also be done using the Rothstein-Trager theorem, see e.g. http://www.math.ubc.ca/~israel/m210/lesson18.pdf ). That doesn't mean there can't be an elementary formula for some definite integrals, but since there doesn't appear to be anything special about $2$ in this context, it's not very likely. The numerical value .35700808127536096106 isn't recognized by Maple's identify or Plouffe's Inverter.
Lets start with some considerations of the indefinite integral:$I=\int \frac{1}{lnx+x+1}dx$ Substitution:$x=e^y$
Therefore integral, I= $\int\frac{e^y}{e^y+y+1}dy$ $=\int\frac{e^y+1}{e^y+y+1}dy-\int\frac{1}{e^y+y+1}dy$ $\approx ln(e^y+y+1)-\int e^{-y} dy $ $\approx ln(e^y+y+1)+e^{-y}$ ----------- (1)
[In the above simplification we have used the information:for large positive values of y $e^y>>y$]
For small positive fractions $e^y \approx 1$ and $e^y>>y$ $ e^y+y+1 \approx 2$ We have, $\int\frac{1}{e^y+y+1}dy \approx \int\frac{1}{2}dy$
Therefore the indefinite integral, I works out to: $I \approx ln(e^y+y+1)-0.5y$ ------------ (2)
For x=1, y=0 For x=2,y=ln2=0.693
Let's do some testing with the second case ie for x=2
The RHS of (2) on differentiation wrt y yields, $\frac{e^y+1}{e^y+y+1}-0.5$ -----(3)
For y=0.693 it produces a value =0.3123
The value of the original integrand for y=.693 is: $ \frac{e^.693}{e^.693+.693+1}=.5415$ We dont have sufficient agreement since the figure 0.693 is not sufficiently good to work out the desired approximation.
For y=0.1 relation (3) gives 0.5022 The original integral produces[for y=0.1] the value:0.50117:we have a good agreement.
But the problem at hand,in consideration of the limits, calls for an improvement.
We proceed as follows:
At the point x=2[ie,y=.693] we approximate the function $e^y$ by a linear element $e^y \approx my+C$ $ e^y=2y+C$
[$m=\frac{d}{dy}(e^y)$. At y=ln2 the derivative is $e^{ln2}=2$]
For y=.693, $e^{0.693}=2$, m=>C=0.614
The Integral:
$\int\frac{1}{e^y+y+1}dy \approx \int\frac{1}{3y+1.614}$ $=0.33ln(3y+1.614)$ We have for the region y=2 the formula for the indefinite integral may approximated as: $ln(e^y+y+1)-0.33ln(3y+1.614)$
It evaluates to:0.87139 for y=2
For y=0 we use relation
$e^y \approx y+1$
$\int\frac{1}{e^y+y+1}dy \approx \int\frac{1}{2y+2}dy$ $=0.5ln(y+1)$ Formula for the indefinite integral= $ln(e^y+y+1)-0.5ln(y+1)$ ---(4) We break up the interval from 1 to 1.6 and then from 1.6 to 2.
The same formula is used for each interval:
For the upper interval we use relation (2) and the lower one we use relation (4)
For the upper interval we obtain:0.87139-0.753=0.11839
For the Lower interval we obtain:0.929-0.693=0.226
The total integral from 1 to 2 is = 0.118+0.226=0.344
A Second Try out:
$\int\frac{e^y}{e^y+y+1}dy=\int\frac{e^y}{e^y+1}\frac{e^y+1}{e^y+y+1}dy$ $=\frac{e^y}{e^y+1}\int\frac{e^y+1}{e^y+1+y}dy-\int\frac{d}{dy}\frac{e^y}{e^y+1}\int\frac{e^y+1}{e^y+y+1}dy$ $\frac{e^y}{e^y+1}ln(e^y+y+1)-\int\frac{e^y}{(e^y+1)^2}ln(e^y+1+y)dy$ We have broken up the indefinite integral into two parts—an accurately determined one and one to be determined in an approximate manner.
Our interval is y=0 to y=ln2
For the second term let’s use a flat value of the integrand corresponding to y=ln1.5
Our formula for the indefinite integral is: $\frac{e^y}{e^y+1}ln(e^y+y+1)-\frac{1.5}{(1.5+1)^2}ln(1.5+1+ln1.5)y+C$
C:Const of Integration.
The same formula is used over the entire integral without breaking it up into subparts.
Definite Integral=
$\frac{2}{2+1}ln(2+ln2+1)- \frac{1.5}{(1.5+1)^2}ln(1.5+1+ln1.5)ln2-0.5ln2 =0.3469788$
This is a better approximation than the previous one and it has been obtained without subdividing the interval.
A STEP FURTHER
We start with the relation:
$I=\int\frac{e^y}{e^y+y+1}dy$ $=\int\frac{e^y}{e^y+1}\frac{e^y+1}{e^y+y+1}dy$ $=\frac{e^y}{e^y+1}ln(e^y+y+1)-\int\frac{e^y}{(e^y+1)^2}ln(e^y+y+1)dy$
$= \frac{e^y}{e^y+1}ln(e^y+y+1)-\int\frac{e^y(e^y+y+1)}{(e^y+1)^3}\frac{ln(e^y+y+1)(e^x+1)}{e^y+y+1}dy$ $= \frac{e^y}{e^y+1}ln(e^y+y+1)- \frac{e^y(e^y+y+1)}{(e^y+1)^3}\frac{[ln(e^y+y+1)]^2}{2}+\int \frac{d}{dt}[\frac{e^y(e^y+y+1)}{(e^y+1)^3}\frac{[ln(e^y+y+1)]^2}{2}$
For the last integral on the RHS of the above step, I am taking a flat value of y=ln1.5 for the integrand. Incidentally, $\frac{d}{dt}[\frac{e^y(e^y+y+1)}{(e^y+1)^3}=\frac{(e^y+1)(2e^{2y}+2e^y+ye^y]-3e^{2y}(e^y+y+1)}{(e^y+1)^4}$ For y=ln1.5, the approximate integral expression reduces to: $\frac{(1.5+1)(2 \times 2.25+2 \times 1.5 +1.5ln1.5)-3 \times 2.25 (1.5+ln1.5+1)}{(1.5+1)^4}[ln(1.5+ln1.5+1)]^2/2y=9.59309 682 \times 10^{-3}y$
Approximate Analytical formula Formula for Antiderivative for the interval [y=0,y=ln2]: $I= \frac{e^y}{e^y+1}ln(e^y+y+1)- \frac{e^y(e^y+y+1)}{(e^y+1)^3}\frac{[ln(e^y+y+1)]^2}{2}+9.59309 682 \times 10^{-3}y$ For y=ln2: $I=\frac{e^{ln2}}{e^{ln2}+1}ln(e^{ln2}+ln2+1)- \frac{e^{ln2}(e^{ln2}+ln2+1)}{(e^{ln2}+1)^3}\frac{[ln(e^{ln2}+ln2+1)]^2}{2}+6.64796117 \times 10^{-3}$ $=0.637947922+0.00664796117$ For y=0, $I=0.286516963$
$\int_{0}^{ln2}\frac{e^y}{e^y+y+1}dy\approx 0.3580$ We are a bit closer this time!
[The accurate part is providing: 0.637947922-0.286516963=0.351430957]
Our basic aim is to determine an approximate analytical expression[a formula] for the antiderivative on some given interval. To this end we express the indefinite integral into an accurately known part [or some accurately known parts]and some other part/s which is/are to be approximated.The overall accuracy of the work would depend on the extent to which the accurately known part dominates over the approximately known component on the interval concerned.
A comparison of the actual Integrand with the approximated integrand obtained by differentiating the approximate formula for the antiderivative for several values on the interval may provide us with the picture of accuracy attained.
$\int_1^2\dfrac{dx}{1+x+\ln x}$
$=\int_0^{\ln2}\dfrac{d(e^x)}{1+e^x+x}$
$=\int_0^{\ln2}\dfrac{e^x}{e^x+x+1}dx$
$=\int_0^{\ln2}\dfrac{1}{1+(x+1)e^{-x}}dx$
$=\int_0^{\ln2}\left(1+\sum\limits_{n=1}^\infty(-1)^n(x+1)^ne^{-nx}\right)dx$
$=\left[x-\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(n-1)!(x+1)^ke^{-nx}}{n^{n-k}k!}\right]_0^{\ln2}$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)
$=\ln2-\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(n-1)!(\ln2+1)^k}{2^nn^{n-k}k!}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(n-1)!}{n^{n-k}k!}$
I would make the substitution $x=a+1$, that one can actually get a power series sans logarithms:
$\int_{1}^{2} \frac{dx}{1+x+\log(x)} = \int_{0}^{1} \frac{da}{2+a+\log(1+a)}$
One may be able to find sequences on oeis that this matches up with, I was not.
(aside: entertainingly the first five terms for the numerators of the power series of $f(a)=\frac{1}{2+a+\log(1+a)}$ match up with oeis:A048607 (Numerators of coefficients in function $a(x)$ such that $a(a(x))$ = $\ln(1+x)$), but the fifth does not, -497 vs -749 for $f(a)$)