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?
 A: Just as User8128 commented, I believe that you are perfectly correct.
Without changing variable and using another CAS
$$\int \mathrm{e}^{-(t-i)^4} \, \mathrm{d}t=\color{red}{-\frac{(t-i)\, \Gamma \left(\frac{1}{4},(t-i)^4\right)}{4 \sqrt[4]{(t-i)^4}}}=-\frac{1}{4} (t-i) E_{\frac{3}{4}}\left((t-i)^4\right)$$
$$\int_{-\infty}^{\infty} \mathrm{e}^{-(t-i)^4} \, \mathrm{d}t=i \left(E_{\frac{3}{4}}(-4)-\Re\left(E_{\frac{3}{4}}(-4)\right)\right)$$ $$\int_{-\infty}^{\infty} \mathrm{e}^{-(t-i)^4} \, \mathrm{d}t=\frac{(-1)^{1/4}\, \Gamma \left(\frac{1}{4},-4\right)}{\sqrt{2}}-i
   \Re\left(-\frac{(-1)^{3/4}\, \Gamma \left(\frac{1}{4},-4\right)}{\sqrt{2}}\right)=2\, \Gamma \left(\frac{5}{4}\right) $$  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.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\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}
