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we know that: $$ \mathscr{E}_1(z)=\int_z^{+\infty}\frac{e^{-t}}{t}\mathrm{d}t,\quad |\arg(z)|<\pi $$ my question is that :

why the $|\mathrm{arg}(z)|<\pi$

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2 Answers 2

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$E_1$ is a multi-branched analytic function; it is conventional to use the negative real axis as a branch cut. The condition $|\arg (z)|<\pi$ ensures that $z$ does not touch the negative real axis, and that the path of integration $t\mapsto z+t$ (for real $0\lt t<\infty$) does not, either.

Unstated in your formula and in the Wikipedia article on the subject, but present in that article's references, is the additional requirement that the path of integration from $z$ to $+\infty$ avoids the negative real axis. Of course one can evaluate the integral for more complicated paths (such as ones that circle the origin several times, say) but the the value of the integral will be different, namely, the value of $E_1$ on some other branch.

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  • $\begingroup$ good,thanks.......... $\endgroup$ Jun 11, 2020 at 13:18
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If you make the change of integration variables $t=s+z$ from $t$ to $s$, you find $$ E_1 (z) = \mathrm{e}^{ - z} \int_0^{ + \infty } \frac{\mathrm{e}^{ - s} }{s + z}\mathrm{d} s. $$ You can see why the negative axis is avoided. It is possible to continue $E_1(z)$ beyond $\arg z =\pm\pi$ as a multivalued analytic function by deforming the contour of integration and taking into account the residue at $s=-z$. It can be shown that $$ E_1 (z) = - \log z - \gamma + \int_0^z {\frac{{1 - \mathrm{e}^{ - t} }}{t}\mathrm{d}t} $$ whenever $z\neq 0$ (here $\gamma$ is the Euler$-$Mascheroni constant). The integral is now an entire function of $z$, whence the multivaluedness is captured in the $\log$.

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  • $\begingroup$ thanks,I mean it $\endgroup$ Jun 11, 2020 at 13:18

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