An integral for which I don't trust WolframAlpha

WolframAlpha tells me that $$\int_{-\infty}^{\infty} e^{-(t-i)^4} \, \mathrm{d}t = 0.$$ I am suspicious of this, because of the idea that for nice enough $f$, since $$\int_{-\infty}^{\infty} f(t-a) \, \mathrm{d}t$$ is holomorphic in $a$ and takes the same value for all $a \in \mathbb{R}$, it should be independent of $a$ by the identity principle; and then $$\int_{-\infty}^{\infty} e^{-(t-i)^4} \, \mathrm{d}t = \int_{-\infty}^{\infty} e^{-t^4} \, \mathrm{d}t = 2 \cdot \Gamma(5/4).$$ But WolframAlpha even tells me numerically that $\int_{-\infty}^{\infty} e^{-(t-i)^4} \, \mathrm{d}t$ is close to $0$. Am I doing something wrong?

• The integrals are equal provided that $i$ is a real number, but it isn't. – WW1 Jun 26 '17 at 3:23
• @WW1 That shouldn't matter by the identity principle, or as User8128 suggests (essentially) the contour of integration can be shifted in the $y$ direction without changing the integral, since it only matters that it connects $\mathrm{Re}[z] = -\infty$ and $\mathrm{Re}[z] = \infty.$ I am pretty confident in either argument but I thought I would ask here just to double check. (I usually trust wolframalpha blindly so it is irritating when something strange like this turns up) – user457218 Jun 26 '17 at 4:06
• @WW1: The argument for $a=i$ isn't about making a change of variable: it's about analytic continuation. – Hurkyl Jun 26 '17 at 10:04
• For the fun type $\int_{-\infty}^{\infty} e^{-(t-\color{red}{1.0}i)^4} \,dt$ and you will get the value !!! – Claude Leibovici Jun 26 '17 at 10:07
• They seem to have already corrected this. – sophros Jun 26 '17 at 17:11

Just as User8128 commented, I believe that you are perfectly correct.

Without changing variable and using another CAS $$\int \mathrm{e}^{-(t-i)^4} \, \mathrm{d}t=\color{red}{-\frac{(t-i)\, \Gamma \left(\frac{1}{4},(t-i)^4\right)}{4 \sqrt{(t-i)^4}}}=-\frac{1}{4} (t-i) E_{\frac{3}{4}}\left((t-i)^4\right)$$ $$\int_{-\infty}^{\infty} \mathrm{e}^{-(t-i)^4} \, \mathrm{d}t=i \left(E_{\frac{3}{4}}(-4)-\Re\left(E_{\frac{3}{4}}(-4)\right)\right)$$ $$\int_{-\infty}^{\infty} \mathrm{e}^{-(t-i)^4} \, \mathrm{d}t=\frac{(-1)^{1/4}\, \Gamma \left(\frac{1}{4},-4\right)}{\sqrt{2}}-i \Re\left(-\frac{(-1)^{3/4}\, \Gamma \left(\frac{1}{4},-4\right)}{\sqrt{2}}\right)=2\, \Gamma \left(\frac{5}{4}\right)$$ Notice that, for the indefinite integral, Wolfram Alpha correctly reports the expression written in red.

What is interesting is that, if you ask Wolfram Alpha $$\int_{0}^{\infty} \mathrm{e}^{-(t-i)^4} \, \mathrm{d}t=-\frac i4E_{\frac{3}{4}}(-1)$$ $$\int^{0}_{-\infty} \mathrm{e}^{-(t-i)^4} \, \mathrm{d}t=+\frac i4E_{\frac{3}{4}}(-1)$$ leading to the $$0$$ you obtained.

There is obviously a bug that I suggest you to report.

• The bug is kind of interesting. Just below i it still works, e.g. for 0.99 i see here at 1 i it fails. – Jannick Jun 26 '17 at 9:36
• @Jannick. Type $1.0i$ and it works !!! – Claude Leibovici Jun 26 '17 at 10:09
• Yes true, that is really funny. For integer one it already fails but for float one it still works. For 1.01 it fails however. – Jannick Jun 26 '17 at 10:13
• @Jannick.**Funny** may not be the most appropriate term here ! Cheers and thanks for your previous comment which made me trying $1.0i$. – Claude Leibovici Jun 26 '17 at 10:15
• @MeniRosenfeld: The original bug does, at the least in Mathematica 9. Integrate[Exp[-(t - I)^4], {t, -Infinity, Infinity}] yields 0. NIntegrate with the same argument returns 1.8128 + very little i, just like what you see. – user22961 Jun 26 '17 at 18:27

$\newcommand{\bbx}{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}{\displaystyle{#1}} \newcommand{\expo}{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}{\mathcal{#1}} \newcommand{\mrm}{\mathrm{#1}} \newcommand{\pars}{\left(\,{#1}\,\right)} \newcommand{\partiald}[]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}{\left\vert\,{#1}\,\right\vert}$ \begin{align} \int_{-\infty}^{\infty}\expo{-\pars{t - \ic}^{4}}\dd t & = \int_{0}^{\infty}\bracks{\expo{-\pars{t - \ic}^{4}} + \expo{-\pars{-t - \ic}^{4}}}\dd t = 2\,\Re\int_{0}^{\infty}\expo{-\pars{t + \ic}^{4}}\dd t = 2\,\Re\int_{\ic}^{\infty + \ic}\expo{-t^{4}}\dd t \\[5mm] & = -2\,\lim_{R \to \infty}\Re\int_{1}^{0}\expo{-\pars{R + \ic y}^{4}}\ic\,\dd y - 2\,\Re\int_{\infty}^{0}\expo{-t^{4}}\dd t - 2\,\Re\int_{0}^{1}\expo{-\pars{\ic y}^{4}}\ic\,\dd y \\[5mm] & = -2\,\lim_{R \to \infty}\Im\int_{0}^{1}\expo{-\pars{R + \ic y}^{4}}\,\dd y + {1 \over 2}\int_{0}^{\infty}t^{-3/4}\expo{-t}\,\dd t \\[5mm] & = -2\,\lim_{R \to \infty}\Im\int_{0}^{1}\exp\pars{-R^{4} - 4R^{3}y\ic + 6R^{2}y^{2} + 4Ry^{3}\ic - y^{4}}\,\dd y + {1 \over 2}\,\Gamma\pars{1 \over 4} \\[1cm] & = {1 \over 2}\,\Gamma\pars{1 \over 4} \\[2mm] & - 2\lim_{R \to \infty}\braces{% \expo{-R^{4}}\int_{0}^{1}\exp\pars{y^{2}\bracks{6R^{2} - y^{2}}} \sin\pars{4Ry\bracks{y^{2} - R^{2}}}\,\dd y} \end{align}

• $\ds{y^{2}\pars{6R^{2} - y^{2}}}$ has a maximum at $\ds{y_{m} = \root{3}R}$. $\ds{y_{m} > 1}$ when $\ds{R > {\root{3} \over 3}}$.
• It vanishes at $\ds{y = 0}$ and at $\ds{y = \root{6}R}$.
• It $\ds{\to -\infty}$ when $\ds{y \to \infty}$.
• It's clear that $\ds{0 < y^{2}\pars{6R^{2} - y^{2}} < 6R^{2} - 1}$ when $\ds{y \in \pars{0,1}}$ with $\ds{R > {\root{3} \over 3}}$.

Then, \begin{align} 0 & < \verts{\expo{-R^{4}}\int_{0}^{1}\exp\pars{y^{2}\bracks{6R^{2} - y^{2}}} \sin\pars{4Ry\bracks{y^{2} - R^{2}}}\,\dd y}_{\ R\ >\ \!\root{3}/3} \\[5mm] & < \expo{-R^{4}}\int_{0}^{1}\exp\pars{1^{2}\bracks{6R^{2} - 1^{2}}}\,\dd y = \exp\pars{-R^{4} + 6R^{2} - 1} \,\,\,\stackrel{\mrm{as}\ R\ \to\ \infty}{\to}\,\,\, \color{#f00}{\large 0} \end{align}

\begin{align} \mbox{such that}\quad \bbx{\int_{-\infty}^{\infty}\expo{-\pars{t - \ic}^{4}}\dd t = {1 \over 2}\,\Gamma\pars{1 \over 4}} \approx 1.8128 \end{align}