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I found this problem by a typo. My homework problem was $\int 2^x \ln(2) \, \mathrm{d}x$ which is $2^x + C$ by the Fundamental Thm of Calculus. I want to be able to solve what I wrote down incorrectly in my homework.

What I wrote for my homework is $\int 2^x \ln(x)\, \mathrm{d}x$ and What I Want to solve, plus I got it wrong. :(

I used integration by parts. $\int u \, \mathrm{d}v = uv - \int v\, \mathrm{d}u$

$\begin{array}{l l} u = \ln(x) & du = \frac{1}{x}\mathrm{d}x \\ \mathrm{d}v = 2^x\mathrm{d}x & v = \frac{2^X}{\ln (2)} \\ \end{array}$

I got this integral:

$\frac{\ln(x)2^x}{\ln 2} - \int \frac{2^x}{x\ln 2}\, \mathrm{d}x$

Another round of integration of parts:

$\begin{array}{l l} u = \frac{2^x}{\ln 2} & du = 2^x\mathrm{d}x \\ \mathrm{d}v = \frac{1}{x}\mathrm{d}x & v = \ln(x) \end{array} $

$\int 2^x \ln(x)\, \mathrm{d}x = \frac{\ln(x)2^x}{\ln 2} - \left[ \frac{2^x \ln x}{\ln 2} - \int \ln(x) 2^x\, \mathrm{d}x \right]$

My final answer is

$ \frac{\ln(x)2^x}{\ln 2} -\frac{2^x \ln x}{\ln 2}= 0$

What did I do wrong?

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    @MaoYiyi Ok. I will2012-11-18

1 Answers 1

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First you did a mistake here: $\int 2^x \ln(x)\, \mathrm{d}x = \frac{\ln(x)2^x}{\ln 2} - \left[ \frac{2^x \ln x}{\ln 2} - \int \ln(x) 2^x\, \mathrm{d}x \right]\Rightarrow \frac{\ln(x)2^x}{\ln 2} -\frac{2^x \ln x}{\ln 2}= 0$ You can't just cancel the integrals, as you will lose the constant of integration. For example $\int \frac{1}{x}\, \mathrm{d}x = \int x^{\prime}\frac{1}{x}\, \mathrm{d}x= x\frac{1}{x} - \int x\frac{-1}{x^2}\, \mathrm{d}x =1+\int \frac{1}{x}\, \mathrm{d}x$ If you cancel the integrals then $1=0$ which is impossible. When canceling integrals one must never forget the constant of integration $c$.

In our case $c=0$. To see this, $\int 2^x \ln(x)\, \mathrm{d}x = \frac{\ln(x)2^x}{\ln 2} - \frac{2^x \ln x}{\ln 2} + \int \ln(x) 2^x\, \mathrm{d}x \Rightarrow 0=\frac{\ln(x)2^x}{\ln 2} - \frac{2^x \ln x}{\ln 2}+c=c$

which leads to $0=0$. Why did this come up? You integrated by parts once and then did the reverse and got back to your starting point. Now how can $\int 2^x \ln(x)\, \mathrm{d}x $ be evaluated? It can't be written as a combination of elementary functions (polynomial,exponential,logarithmic,trigonometric and hyperbolic functions and their inverses). I will show this for $\int e^x \ln(x)\, \mathrm{d}x $.

$\int e^x \ln(x)\, \mathrm{d}x = \int (e^x)^{\prime} \ln(x)\, \mathrm{d}x=e^x\ln x- \int e^x (\ln(x))^{\prime}\, \mathrm{d}x=e^x\ln x-\int \frac{e^x}{x}\, \mathrm{d}x $ The last integral is not elementary as shown by Risch Algorithm. For more information look here: Exponential Integral. And no, I don't think there is any book covering this topic

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    if you write one, i'll download it.2012-11-22