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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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Well, nothing...but you just proved that $\,0=0\,$ , which seems to be a result pretty far away from spectacular, and way m ore important: you did not solve what you were asked! –  DonAntonio Nov 18 '12 at 12:50
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Are you trying to find $\int 2^x\ln(2)dx$ or $\int 2^x\ln(x)dx$? You mention both. –  yunone Nov 18 '12 at 12:52
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Your $du$ and $dv$ should have $dx$ at the right. After that you reversed simply the integration by parts and are back at your starting point... –  Raymond Manzoni Nov 18 '12 at 12:52
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The integral of $2^x\ln(x)$ is not elementary so you can't compute its antiderivative. If it were then the integral of $\frac{2^x}{x}$ and so $\frac{e^x}{x}$ would have to be elementary. See en.wikipedia.org/wiki/Exponential_integral –  Nameless Nov 18 '12 at 12:52
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@DonAntonio but the typo taught me more! –  yiyi Nov 18 '12 at 13:08
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1 Answer 1

up vote 2 down vote accepted

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. –  yiyi Nov 22 '12 at 0:46
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