Solved ... with FOLLOW-UP!
Hello, I have progressed and found a workable solution.
1) Following
HallsofIvy's help, I scoured the internet looking for further assistance on integrating \(\displaystyle \int \exp(x) \ \ln(x) \, dx \).
2) I found the following
LINKED German forum site, uploading it with the German translated into English. The discussion here centered on a student trying to get his work (on integrating the above) to match his teacher's; the teacher's solution was also given.
3) In this thread's discussion,
DrDirectX, recommended searching for keyword:
Integral Exponential Function. And,
Dr_Sonnhard_Graubner, mentioned going to the site "wolfram".
4) Then, with the help of yet another forum site (can't recall it), which also suggested the wolfram site, I ended up at
http://integrals.wolfram.com/index.jsp.
What an amazing site!!! I am sure you guys are aware of this site.
NOTE: For anyone needing quick solutions on integrals, this site is very helpful.
5) This site gave the solution for \(\displaystyle \int \exp(x) \ \ln(x) \, dx \text{ ... as } = \exp(x) \ \ln(x) - E_i(x) \text{ where … } E_i(x) \text{ is the Exponential Integral Func}\). ... I had to read up on \(\displaystyle E_i(x)\), which was new to me. There are plenty of internet resources to help on that.
6) Then, I re-checked the wolfram site for the solution to the more specific: \(\displaystyle \int \exp(x/k) \ \ln(1 + bx) \, dx \text{ where … } b = r/C_1 \). Amazingly, it provided one!
7) The wolfram.com solution for this is as follows: (using the above specific formula constants) ...
... Solution to \(\displaystyle \int \exp(t/k) \ \ln(1 + bt) \, dt = k \exp(t/k) \ \ln(\dfrac{C_2}{C_1}) - k \exp(\dfrac{-C_1}{rk}) \Big[ E_i(C_2/rk) - E_i(C_1/rk) \Big] \text{ where … }C_2 = C_1 + rt\)
...
NOTE: This solution includes the constant term (@ t=0) which is ... \(\displaystyle = k \exp(\dfrac{-C_1}{rk}) \ E_i(C_1/rk)\)
8) Combining this integral solution with the rest of the equation, results in: ... Note: The stuff inside the \(\displaystyle \Big\{ \ \Big\}\) is the integral solution.
... \(\displaystyle \exp(t/k) \ T = (a/k) \left\{ k \exp(t/k) \ \ln(\dfrac{C_2}{C_1}) - k \exp(\dfrac{-C_1}{rk}) \Big[ E_i(C_2/rk) - E_i(C_1/rk) \Big] \right\}\)
9) Further simplifying results in the
final equation: \(\displaystyle T = a \left\{\ln(\dfrac{C_2}{C_1}) - \dfrac{\left[\ln(\dfrac{C_2}{C_1})\ + \displaystyle \sum_{n=1}^{\infty} \left(\dfrac{(C_2/rk)^n - (C_1/rk)^n}{n \ n!} \right)\right]}{\exp(\dfrac{C_2}{rk})}\right\} \)
... Here, the \(\displaystyle E_i(x) \text{ terms}\) have been reduced to their most basic series expressions. This equation can be used in an Excel worksheet using array functions (i.e. SumProduct(array(1), array(2), etc)) to solve for the series solutions.
Executing the SERIES functions from n=1 to n=12 yields plenty of resolution to the final answer. Variable definitions are as follows: ...
\(\displaystyle a = 3.0 \text{ deg-C} / \ln(2)\)
\(\displaystyle k = 50 \text{ years}\)
\(\displaystyle r = 2.2 \text{ ppm/yr ... Note: This was lower prior to 2000. A more advanced approach would account for a changing input here.}\)
\(\displaystyle C_1 = 310 \text{ ppm in 1955,}\)
\(\displaystyle t = \text{ # years since in 1955, and}\)
\(\displaystyle C_2 = \text{ ppm in desired year} = C_1 + r t \text{ , and}\)
\(\displaystyle T = \text{ temperature for the desired year}\)
LINKED HERE is an Excel worksheet (ver2010) that contains the final results. It uses both this model formula (see Temp2) plus a different, simpler one (Temp1) that actually gives better results. The calculated results are very interesting. ... Thanks for the help, sauerj
