Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Can you help me find the value of the integral

$$\int_{1}^{2}\frac{dx}{1+x+\ln x}$$

Thank you

share|improve this question
7  
wolframalpha.com/input/… –  Matt N. Dec 29 '11 at 12:32
    
@Ismail: Do you want an answer or a method for evaluating the given integral? –  user9413 Dec 29 '11 at 13:26
    
write $\log$ as a power series and divide, use a few terms for an approximation –  yoyo Dec 29 '11 at 13:55
    
@Chandrasekhar : I would like a method more than an answer. Teach me how to fish and give me no fish. –  user20010 Dec 29 '11 at 20:21
1  
Do you have reason to believe there is a nice closed-form expression? Most integrals don't have one, so you are left with numerical methods. –  Nate Eldredge Jan 2 '12 at 18:23

4 Answers 4

up vote 6 down vote accepted

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.

share|improve this answer

$\int_1^2\dfrac{dx}{1+x+\ln x}$

$=\int_0^{\ln2}\dfrac{d(e^x)}{1+e^x+x}$

$=\int_0^{\ln2}\dfrac{e^x}{e^x+x+1}dx$

$=\int_0^{\ln2}\dfrac{1}{1+(x+1)e^{-x}}dx$

$=\int_0^{\ln2}\left(1+\sum\limits_{n=1}^\infty(-1)^n(x+1)^ne^{-nx}\right)dx$

$=\left[x-\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(n-1)!(x+1)^ke^{-nx}}{n^{n-k}k!}\right]_0^{\ln2}$ (can be obtained from http://en.wikipedia.org/wiki/List_of_integrals_of_exponential_functions)

$=\ln2-\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(n-1)!(\ln2+1)^k}{2^nn^{n-k}k!}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n(n-1)!}{n^{n-k}k!}$

share|improve this answer

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 $f(a)$)

share|improve this answer

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.

share|improve this answer
    
Relation (2) will give a better approximation for smaller values of y . You may try y=0.1 or 0.01 to test the situation. Relation (1) seems to work well with y=5 . –  Anamitra Palit Apr 21 '12 at 6:57
    
$\int_{5}^{9}\frac{1}{x+lnx+1}dx=\int_{ln5}^{ln9}\frac{e^y}{e^y+y+1}dy=0.4784$--‌​(A) Formula used:Indefinite Int=$ln(e^y+y+1)+e^y$--(B). Mathematical software is giving:0.410145.If the RHS of B is differentiated and arbitrary values from 5 to 9 are inserted in the resulting relation there is some agreement with the original integrand for the inserted values.But the interval from 1 to 2 is a tricky one.It would be difficult to find a single relation to cover the whole integral adequately.So we have to break up the interval providing each interval with a formula of its own. –  Anamitra Palit Apr 21 '12 at 11:31
    
For large values of x such that $lnx>5$ formula (B) should work adequately. For small values of y=lnx ie small +ve fractions,we may use the formula:$ln(e^y+y+1)+0.5e^{-y}$ . For intervals of the type ,y=1 to y=2,we should use the least number of sub-intervals to produce the desired accuracy,taking care that the the same formula is used for each subpart/subinterval –  Anamitra Palit Apr 21 '12 at 11:42
    
A CRUDE ERROR ESTIMATE[Absolute Error]: Maximum Inaccuracy Induced by the Approximated integral < Maximum value of the integrand in the approximated integral $$\times$$ Interval Length –  Anamitra Palit Apr 25 '12 at 12:41
1  
What's the point of all this? The integrand being nice and analytic in the interval $[1,2]$, all the standard numerical methods should work well to give good numerical approximations. But approximations can't give you an exact closed-form formula. –  Robert Israel Apr 25 '12 at 16:18

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.