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For real nonzero values of x, the exponential integral $\;\mbox{Ei}(x)\;$ may be defined as: $$ \mbox{Ei}(x) = \int_{-\infty}^{x} \frac{e^t}{t}\:dt $$ I have more than one reason to believe in the following conjecture : $$ \int_{-\infty}^{+\infty} \mbox{Ei}(-x^2/2)\:dx = -2\sqrt{2\pi} $$ But could only "prove" this result numerically. Can someone provide a "real" proof?

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3 Answers 3

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$$ \int_{-\infty}^{+\infty} \mbox{Ei}(-x^2/2)\:\mathrm{d}x = 2 \int_{0}^{+\infty} \mbox{Ei}(-x^2/2)\:\mathrm{d}x = -2 \int_{0}^{+\infty} \int_{x^2/2}^\infty \tfrac{1}{t} \mathrm{e}^{-t}\:\mathrm{d}t\:\mathrm{d}x $$ Since the region of integration is $\left\{(t,x) \colon 2 t > x^2 \right\} = \left\{(t,x) \colon x < \sqrt{2t} \right\}$, interchanging integration order, as warranted by Tonelli theorem: $$ -2 \int_{0}^{+\infty} \left(\int_{x^2/2}^\infty \tfrac{1}{t} \mathrm{e}^{-t}\:\mathrm{d}t\right)\:\mathrm{d}x = -2 \int_0^\infty \left( \int_0^{\sqrt{2 t}} 1 \: \mathrm{d}x\right) \tfrac{1}{t} \mathrm{e}^{-t} \: \mathrm{d}t = -2 \int_0^\infty \sqrt{\frac{2}{t}} \mathrm{e}^{-t} \:\mathrm{d}t $$ Now with the change of variables $t = u^2/2$: $$ -2 \int_0^\infty \sqrt{\frac{2}{t}} \mathrm{e}^{-t} \:\mathrm{d}t = -4 \int_0^\infty \mathrm{e}^{-u^2/2} \:\mathrm{d}u = -4 \sqrt{\frac{\pi}{2}} = -2 \sqrt{2 \pi} $$

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  • $\begingroup$ An answer with (sign) errors cannot be accepted. Would you please correct them. $\endgroup$ Apr 6, 2014 at 12:18
  • $\begingroup$ @HandeBruijn Fixed. Thanks for pointing those out. $\endgroup$
    – Sasha
    Apr 6, 2014 at 13:54
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$$ \int_{-\infty}^{+\infty} \mbox{Ei}(-x^2/2)~dx=2\int_{0}^{+\infty} \mbox{Ei}(-x^2/2)~dx\\ =2\left(\left.x\mbox{Ei}(-x^2/2)\right|_{0}^{+\infty}-\int_{0}^{+\infty} x~d\mbox{Ei}(-x^2/2)\right)\\ =-2\int_{0}^{+\infty} x~d\mbox{Ei}(-x^2/2)\\ =-4\int_{0}^{+\infty} e^{-x^2/2}dx =-4\sqrt{\pi/2}=-2\sqrt{2\pi}. $$

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Denote the considered integral by $I$. Let $\Omega=\left\{(t,x):t+\frac{x^2}{2}<0\right\}$. Then clearly we have $$ I=\int_{-\infty}^\infty\left(\int_{-\infty}^\infty \frac{e^t}{t}\chi_{\Omega}(t,x)dt\right)dx. $$ Now the integrand satisfies $f(t,x)=\dfrac{e^t}{t}\chi_{\Omega}(t,x)\leq0$ for $(t,x)\in\mathbb{R}$, so we can use Tonelli's theorem, (applied to $-f$ if you wish,) to interchange the order of integration. Thus $$\eqalignno{ I&=\int_{-\infty}^\infty\frac{e^t}{t}\left(\int_{-\infty}^\infty \chi_{\Omega}(t,x)dx\right)dt\cr &=\int_{-\infty}^0\frac{e^t}{t}\left(\int_{-\sqrt{-2t}}^{\sqrt{-2t}}dx\right)dt\cr &=2\int_{-\infty}^0\frac{e^t}{t}\sqrt{-2t}\,dt&(t\leftarrow-t)\cr &=-2\sqrt{2}\int_{0}^{\infty}\frac{e^{-t}}{\sqrt{t}}\,dt\cr &=-2\sqrt{2}\Gamma\left(\frac{1}{2}\right)=-2\sqrt{2\pi} } $$ which is the desired answer.

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