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Evaluate the integral $$\int_1^\infty \frac{2^x}{2^{(2^x)}}dx$$

My Try:

substituting $t = 2^x$ we get:

$$\ln 2 \int_2^\infty \left(\frac{1}{2}\right)^t dt = \frac{\ln 2}{\ln 0.5} \left( \left(\frac{1}{2}\right)^\infty - \left(\frac{1}{2}\right)^2 \right)$$

Apparently $\frac{\ln 2}{\ln 0.5} = -1$ so we get that the integral equals $\frac{1}{4}$.

But that's a false proof.

Where is my mistake and how to correct that?

Thanks.

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  • $\begingroup$ What different it would make? $(1/2)^n \to 0$. $\endgroup$ – Elimination Jan 30 '15 at 15:21
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    $\begingroup$ $ln(2)$ goes into denominator $\endgroup$ – Arashium Jan 30 '15 at 15:24
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$t=2^x$,$x=\frac{\ln t}{\ln 2}$,thus $dx=\frac{1}{\ln2}\frac1tdt$, I'm afraid the integral should be $$\int_1^\infty \frac{2^x}{2^{(2^x)}}dx=\frac{1}{\ln 2} \int_2^\infty \left(\frac{1}{2}\right)^t dt $$

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  • $\begingroup$ Yes, and the final answer is $\frac{1}{ln(2) ln(0.5)}\times (0-\frac{1}{4})$ $\endgroup$ – Arashium Jan 30 '15 at 15:27
  • $\begingroup$ @Arashium Thanks. But the rest is plain to see. $\endgroup$ – Vim Jan 30 '15 at 15:28
  • $\begingroup$ Just for his verification :) $\endgroup$ – Arashium Jan 30 '15 at 15:29
  • $\begingroup$ Hey @vim, thank you. I hope that's not rude but would you mine looking at another answer of mine? math.stackexchange.com/q/1126519/160028 $\endgroup$ – Elimination Jan 30 '15 at 15:29
  • $\begingroup$ Hello @Elimination , I'm sorry but I have just taken one semester's course in analysis and this problem is quite beyond me. $\endgroup$ – Vim Jan 30 '15 at 15:32

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