Although this question is going to seem completely trivial to anyone with any exposure to path integrals, I'm looking to answer this precisely and haven't been able to find any materials after looking for about 40 minutes, which leads me to believe that it makes sense to ask the question here. In particular I'm looking for an answer wherin any mathematical claims are phrased as precisely as possible, with detailed proofs either provided or referenced. Also my search for a solution has led me to think I'm actually looking for a good explenation of Wick rotation, which I can't really claim to understand in detail. Any good references about this would be very welcome as well.

I'm looking to make sense of the following integral identity:

$$\int_{-\infty}^{\infty} dx \ \exp\left(i\frac{a}{2}x^2+iJx\right)=\left(\frac{2\pi i}{a}\right)^{1/2}\exp\left(\frac{-iJ^2}{2a}\right), \qquad a,J\in\mathbb{R}$$

Wikipedia (and various other sources) say that "This result is valid as an integration in the complex plane as long as a has a positive imaginary part." Clearly the left hand side does not exist in Lebesgue sense for real $a, J$. An answer to the question "Wick rotation in field theory - rigorous justification?" claims:

"it is convergent as a Riemann integral, thanks to some rather delicate cancellations. To make the integral well defined -- equivalently to see how these cancellations occur -- we need to supply some additional information. Wick rotation provides a way of doing this. You observe that the left hand side is analytic in t , and that the right hand side is well-defined if Im(t)<0. Then you can define the integral for real t by saying that it's analytic continued from complex t with negative imaginary part."

I want to see the gory details and all known motivation for the validity of this procedure for the kinds of applications where such integrals occur. Suggestions such as "include an $i\epsilon$ in order to make it finite" seem arbitrary. In that case one would have to motivate that prescription from the very start, that is within the modeling procedure that ends up giving that integral expression (which is likely the correct way to approach this). I'm also not sure how to interpret the right hand side, since it involves the square root of an imaginary number, which should involve some choice of branch cut, which I have never seen specified in connection to this formula.

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    $\begingroup$ Probably trivial. Doesn't forming a closed contour : (i) Real axis, (ii) $e^{i\pi/4}*Real axis and (iii) Quarter circular arcs at infinity. Followed by analyticity arguments leading to familiar real integral is rigorous enough. $\endgroup$
    – Sunyam
    Nov 11, 2017 at 9:30
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    $\begingroup$ This article by D. Desbrow might help: American Mathematical Monthly 105.8 (1998), pp. 726–731 [ JSTOR]. $\endgroup$
    – AlQuemist
    Mar 8, 2018 at 8:46

2 Answers 2


Proposition. Let there be given two complex numbers $a,b\in \mathbb{C}$ such that ${\rm Re}(a)\geq 0$. In the case ${\rm Re}(a)=0$, we demand furthermore that ${\rm Im}(a)\neq 0$ and ${\rm Re}(b)=0$. The Gaussian integral is well-defined and is given by $$ \underbrace{\int_{\mathbb{R}}\!dx~ e^{-\frac{a}{2}x^2+bx}}_{=:~ I_{\mathbb{R}}(a,b)} ~=~\lim_{\begin{array}{c} x_i\to -\infty \cr x_f\to \infty \end{array} } \underbrace{\int_{[x_i,x_f]}\!dx~ e^{-\frac{a}{2}x^2+bx}}_{=:~ I_{[x_i,x_f]}(a,b)} ~=~\underbrace{\sqrt{\frac{2\pi}{a}}e^{\frac{b^2}{2a}}}_{=:~ F(a,b)}, \tag{A}$$ where it is implicitly understood that the square root has positive real part.

Remark: The Riemann/Darboux integral is not defined for non-bounded sets, so it can only be used for the middle expression of eq. (A).

I) Sketched proof in case of ${\rm Re}(a)> 0$: The function $g(x)=e^{-\frac{{\rm Re}(a)}{2}x^2+{\rm Re}(b)x}$ serves as a majorant function for Lebesgue's dominated convergence theorem, which establishes the first equality of eq. (A). For the second equality of eq. (A), we divide the proof into cases:

  1. Case $a>0$ and $b\in \mathbb{R}$. Complete the square. $\Box$

  2. Case $a>0$. Complete the square. Shift the integration contour appropriately to a horizontal line in the complex plane in order to reduce to case 1, cf. Cauchy's integral theorem. Argue that contributions at infinity vanish. $\Box$

  3. Case ${\rm Re}(a)> 0$. Rotate the integration contour to a line of steepest descent in order to reduce to case 2, cf. Cauchy's integral theorem. Argue that contributions at infinity vanish. $\Box$

II) Sketched proof in the oscillatory case ${\rm Re}(a)=0, {\rm Im}(a)\neq 0, {\rm Re}(b)=0$: The lhs. of eq. (A) is not Lebesgue integrable. It is an improper integral defined via the middle expression of eq. (A). It remains to prove the second equality of eq. (A). It is possible to give a proof using Cauchy's integral theorem along the lines of Jack's answer. In this answer we will instead give a proof in the spirit of an infinitesimal deformation prescription.

Given $\varepsilon>0$. As $x_i\to \infty$ and $x_f\to \infty$ it is not hard to see that $I_{[x_i,x_f]}(a,b)$ oscillates with smaller and smaller amplitude that tends to zero, and it is hence convergent without any regularization. The convergence improves if we let $a$ have a positive real part. In other words, the convergence is uniform wrt. ${\rm Re}(a)\geq 0$, i.e.

$$ \exists X_i,X_f\in \mathbb{R} ~\forall x_i\leq X_i~\forall x_f\geq X_f ~\forall {\rm Re}(a)\geq 0:~~ \left| I_{[x_i,x_f]}(a,b)- I_{\mathbb{R}}(a,b)\right| ~\leq~\frac{\varepsilon}{4}.\tag{B}$$

Next use Lebesgue's dominated convergence theorem with majorant function of the form $g(x)=C~1_{[x_i,x_f]}(x)$ (where $C>0$ is an appropriate constant) to argue that

$$I_{[x_i,x_f]}( i{\rm Im}(a),b) ~=~\lim_{{\rm Re}(a)\to 0^+} I_{[x_i,x_f]}(a,b) , \tag{C}$$

i.e. $\exists {\rm Re}(a)>0$ such that

$$ \left| I_{[x_i,x_f]}( i{\rm Im}(a),b)-I_{[x_i,x_f]}( a ,b) \right| ~\leq~\frac{\varepsilon}{4},\tag{D}$$


$$\left| \underbrace{F( a ,b)}_{=~ I_{\mathbb{R}}(a,b)}- F( i{\rm Im}(a),b) \right| ~\leq~\frac{\varepsilon}{4}.\tag{E}$$

In eq. (E) we used that the function $F$ is continuous. All together, eqs. (B), (D) & (E) yield $$ \begin{align} \left| I_{\mathbb{R}}(i{\rm Im}(a),b) - F( i{\rm Im}(a),b)\right| ~\leq~&\left| I_{\mathbb{R}}(i{\rm Im}(a),b)- I_{[x_i,x_f]}( i{\rm Im}(a),b)\right|\cr &+\left| I_{[x_i,x_f]}( i{\rm Im}(a),b) - I_{[x_i,x_f]}( a ,b)\right| \cr &+\left| I_{[x_i,x_f]}(a,b)- I_{\mathbb{R}}(a,b)\right|\cr &+\left|F( a ,b) - F( i{\rm Im}(a),b)\right| \cr ~\leq~&\varepsilon.\end{align} \tag{F}$$

Eq. (F) shows that the second equality of eq. (A) holds. $\Box$

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    $\begingroup$ Thank you for this answer! Could you elaborate a bit on what you mean in sketch I, case 3, when you say "Rotate the integration contour to a line of steepest descent"? Or perhaps point me to somewhere I might read more about it? $\endgroup$ Feb 27, 2019 at 14:53

enter image description here

First of all, one need to prove the following key integral formula: \begin{align*} \boxed{I = \int_{-\infty}^{+\infty} d x e^{ i a x^2} = \sqrt{\dfrac{i \pi}{a}} \qquad (a>0) } \end{align*} Usually one can pick up an analytic function $f(z)=e^{ i a z^2}$ and then performs the complex integral along the closed contour showed above. $$J = \oint d z e^{ i a z^2} = \int_{C_1} + \int_{C_2} + \int_{C_3} + \int_{C_4} = J_1 + J_2 + J_3 + J_4 =0$$ Next one need calculate the four line integrals along four different intervals.

  1. $C_1:z=x \qquad x \in [-p,p]$ $$J_1 = \int_{C_1} dz e^{i a z^2} = \int_{-p}^p d x e^{i a x^2}$$
  2. $C_2:z=p+iy \qquad y \in [0,p]$ $$J_2 = \int_{C_2} dz e^{i a z^2} = \int_0^p i d y e^{i a (p+i y)^2}$$
  3. $C_3:z=x+iy = (1+i) y = \sqrt{2}e^{i\pi/4}y \qquad y \in [p,-p]$ \begin{align*} J_3 & = \int_{C_3} dz e^{i a z^2} = e^{i\pi/4} \int_{\sqrt{2}p}^{-\sqrt{2}p} d x e^{- a x^2} \end{align*}
  4. $C_4:z=-p + i y \qquad y \in [-p,0]$ \begin{align*} J_4 & = \int_{C_4} dz e^{i a z^2} = \int_{-p}^0 i d y e^{i a (i y -p)^2} \qquad (y \Rightarrow -y) \\ & = \int_p^0 - i d y e^{i a (-i y - p)^2} = \int_0^p i d y e^{i a (i y+p)^2} \\ & = J_2 \\ \end{align*}

In the final step, we will consider the limiting process:$p\rightarrow+\infty$ \begin{align*} & 0 \leq |J_2| \leq \left|\int_0^p i d y e^{i a (i y+p)^2}\right| \leq \int_0^p d y e^{-2 a p y} = \dfrac{1-e^{-2a p^2}}{2 a p} \\ & \lim_{p \rightarrow +\infty} \dfrac{1-e^{-2a p^2}}{2 a p} \Rightarrow \lim_{p\rightarrow+\infty} |J_2| = 0 \Rightarrow \boxed{\lim_{p\rightarrow +\infty} J_2 = \lim_{p\rightarrow +\infty} J_4 =0} \end{align*} \begin{align*} & p\rightarrow +\infty \Rightarrow J_1+J_3=0 \Rightarrow J_1 = -J_3 = e^{i\pi/4} \int_{-\infty}^{+\infty} d x e^{- a x^2} = e^{i \pi/4} \sqrt{\dfrac{\pi}{a}} = \sqrt{\dfrac{i\pi}{a}} \\ & \Rightarrow \boxed{\int_{-\infty}^{+\infty} d x e^{i a x^2} = \sqrt{\dfrac{i\pi}{a}} \qquad (a>0)} \end{align*} One can also derive a similar result for $a<0$ with the help of the following closed contour. enter image description here

Then for the integral you are giving: \begin{align} \int_{-\infty}^{+\infty} e^{\dfrac{iax^2}{2}+iJx} dx = \int_{-\infty}^{+\infty} e^{i \dfrac{a}{2}(x+\dfrac{J}{a})^2-i\dfrac{J^2}{2a}} dx = \sqrt{\dfrac{i2\pi}{a}}e^{-i\dfrac{J^2}{2a}} \end{align}

Hope it helps.

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    $\begingroup$ Thank you very much for your answer. I will gladly accept it but it leaves some questions about the precise mathematical statements meant. I still don't know why the square root is being used to apply to imaginary arguments (apparantly as an abbreviation of $e^{i\pi/4} $. ) While the contour method in it's rigorous form is described in texts on complex analysis, I would still like a precise mathematical statement of the statement being proved. In particular one of my main questions was "in what sense does the integral exist", since it clearly doesn't in Lebesgue sense. $\endgroup$ Mar 8, 2018 at 13:30
  • $\begingroup$ This integral exists guaranteed by Cauchy's theorem in complex analysis. For connection to more advanced theory, I cannot talk more. $\endgroup$
    – Jack
    Mar 8, 2018 at 15:21
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    $\begingroup$ @Jack : I think results like $\sqrt{i \frac{\pi}{a}}$ are ambiguous. Which branch of the square-root function shall one take? $\endgroup$
    – AlQuemist
    Mar 12, 2018 at 12:15

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