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I am trying to find the expected value of a probability density function. Solving the integral of the function times its random variable with integration by parts, I arrive at the following integrals which are rather complex. I'd appreciate very much if you could provide some directions on how I could tackle this problem. Thanks!

$$\int_0^\infty \ln(2x+1) \, x e^{-2Tx} \, dx + \int_0^\infty \ln(2x+1) \, e^{-2Tx} \, dx$$

Alex M.
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user283870
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  • I edited your expression into proper LaTeX form. Please make sure that it is still what you want to solve. Also, it would be helpful if you could give the original problem that you are trying to solve as you might be able to avoid these integrals altogether. :) – Rubarb Oct 25 '15 at 22:58
  • @Rubarb : You wrote $ln(2x+1)$ instead of $\ln(2x+1)$, coded as \ln(2x+1). The backslash not only prevents italiciziation, but also results in proper spacing to the left and right in expressions like $a\ln b$ and to the left in expressions like $a\ln(b)$. It is standard usage. The same applies to \sin, \cos, \max, \sup, \det, \log, \exp, etc. ${}\qquad{}$ – Michael Hardy Oct 25 '15 at 23:07
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    Actually, user283870 wrote that. I did minimal edits to make it make some sense. (For example, exponents were on the ground due to lack of brackets, etc...) – Rubarb Oct 25 '15 at 23:31
  • Where is the logarithm in this? Ok, I edited the title, since the edited question doesn't contain logarithms – Yuriy S Aug 07 '16 at 16:31

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Sub $x=(u-1)/(2 N)$; then the integral is equal to

$$\frac1{2 N} e^{T/N} \int_1^{\infty} \frac{du}{u} e^{-T u/N} = \frac1{2 N} e^{T/N} \Gamma \left (0,\frac{T}{N} \right )$$

where $\Gamma$ is the upper incomplete gamma function.

Ron Gordon
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