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How to prove \begin{equation} \int_{-b}^{\infty}\log^{\nu}(t+b)e^{-t}e^{-e^{-t}}dt\xrightarrow{b\rightarrow\infty} \log^{\nu}(b), \end{equation} where $\nu$ is any real or complex number. I have validated this asymptotic in MATLAB with numerical method.

Note that in our previous post (please refer to How to approximate the integral $\int_{-b}^{\infty}\log(t+b)e^{-t}e^{-e^{-t}}dt$), the special case $\nu=1$ was proved by @tired, however, this method only applies to the case $\nu=1$ rather than the general value of $\nu$. There should be a general method to prove it.

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kawofengche
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  • how do you define $\log^{\nu}(z)$ on the interval $z \in {0,1}$? – tired Nov 14 '16 at 18:53
  • sorry, I didn't get your concerning... – kawofengche Nov 14 '16 at 21:11
  • for $t+b<1$ the logarithm gets negative. Now we have to answer the question how to take the non-integer exponent of this quantity (you may recall the for example the square root is a multivalued function in the complex plane) – tired Nov 14 '16 at 21:14
  • Take the definition in MATLAB: For negative and complex numbers z = u + iw, the complex square root sqrt(z) returns sqrt(r)(cos(phi/2) + 1i*sin(phi/2)) where r = abs(z) is the radius and phi = angle(z) is the phase angle on the closed interval -pi <= phi <= pi. – kawofengche Nov 14 '16 at 21:31
  • Suppose that $\int_{-b}^{\infty}\log^{\nu}(t+b)e^{-t}e^{-e^{-t}}dt\xrightarrow{b\rightarrow\infty} \log^{\nu}(b)$ holds, if we can prove $\int_{-b}^{\infty}\log^{\nu+\delta}(t+b)e^{-t}e^{-e^{-t}}dt\xrightarrow{b\rightarrow\infty} \log^{\nu+\delta}(b)$, then it accomplish the proof. Is it simpler to prove in such way? – kawofengche Nov 15 '16 at 10:47

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