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I'm trying to evaluate a difficult integral. I'm able to break it down in separate terms and deal with scalar multiplication. However, I'm stuck trying to evaluate two terms in particular. Here is the first one:

$$ \int_0^y \exp\left[-\frac{2^x + 2^{y-x}-2}{a}\right]dx $$

And the second one (there is no error in the signs of the second factor):

$$ \int_0^y \exp\left[-\frac{2^x + 2^{y-x}-2}{a}\right]\left(\frac{2^{-x}+2^{x-y}}{a}\right)dx $$

Can anyone help me solve these?

EDIT:

Here is some more information on my problem. Basically, I'm trying to find the PDF associated to the sum of two i.i.d. RVs: $\underline{Y} = {\underline{X}} + \underline{X}$. I know I can get it by using the convolution, i.e., $f_{\underline{Y}}(y) = (f_{\underline{X}} * f_{\underline{X}})(y) = \int_{-\infty}^{\infty} f_{\underline{X}}(\tau) f_{\underline{X}}(y-\tau)d\tau$.

After a few steps and a few substitutions, I'm stuck at trying to evaluate the two integrals in my original question (there are in fact other remainders from this integration, but I can deal with everything else other than those two integrals). Below is the equation for $f_{\underline{X}}(x)$, where $a = \frac{2\lambda^2P}{\sigma^2}$ is only a constant term.

$$ f_{\underline{X}}(x) = \begin{cases}\ln (2) \exp\left[-\frac{(2^x-1)\sigma^2}{2\lambda^2P}\right]\left(2^{-x}+\frac{\sigma^2}{2\lambda^2P}\right) &, x \geq 0\\0 &, \textrm{otherwise}\end{cases} $$

Also, if that helps, I'm actually not interested in the PDF of ${\underline{Y}}$, but only its CDF. Therefore, I need to take the integral in $y$ from 0 to some value $\hat{y}$ of the result:

$$ F_{\underline{Y}}(\hat{y}) = \int_0^{\hat{y}} f_{\underline{Y}}(y)dy $$

I don't think I can switch the two integrals however or do some clever trick, but then again, I might be mistaken.

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  • $\begingroup$ For general $y,$ it won't have a simple form. If $y~0$ or $y \to \infty$ then it's possible something interesting might result. Explain what your needs are for these integrals. $\endgroup$
    – user321120
    Commented Oct 12, 2020 at 22:14
  • $\begingroup$ Hi @skbmoore, thanks for looking into this problem. I added some more explanation in my original question. I hope this helps. $\endgroup$
    – Olivier
    Commented Oct 14, 2020 at 0:25

1 Answer 1

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For $\int_0^ye^{-\frac{2^x+2^{y-x}-2}{a}}~dx$ ,

$\int_0^ye^{-\frac{2^x+2^{y-x}-2}{a}}~dx$

$=e^\frac{2}{a}\int_0^ye^{-\frac{2^x+2^{y-x}}{a}}~dx$

$=\int_0^y\sum\limits_{n=0}^\infty\dfrac{(-1)^ne^\frac{2}{a}(2^x+2^{y-x})^n}{a^nn!}~dx$

$=\int_0^y\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^ne^\frac{2}{a}2^{(n-k)x}2^{k(y-x)}}{a^nn!}~dx$

$=\int_0^y\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^ne^\frac{2}{a}2^{ky}2^{(n-2k)x}}{a^nk!(n-k)!}~dx$

$=\int_0^y\left(e^\frac{2}{a}+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^ne^\frac{2}{a}2^{ky}2^{(n-2k)x}}{a^nk!(n-k)!}\right)~dx$

$=\left[e^\frac{2}{a}x+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^ne^\frac{2}{a}2^{ky}2^{(n-2k)x}}{a^nk!(n-k)!(n-2k)\ln2}\right]_0^y$

$=e^\frac{2}{a}y+\sum\limits_{n=1}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^ne^\frac{2}{a}(2^{(n-k)y}-2^{ky})}{a^nk!(n-k)!(n-2k)\ln2}$

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  • $\begingroup$ Thanks, this is great! I'll try using a similar approach for the other integral and I'll see how it goes $\endgroup$
    – Olivier
    Commented Oct 14, 2020 at 12:47

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