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$$\int_{-\infty}^\infty \text{e}^{\ a\ (x+b)^2}\ \text dx=\int_{-\infty}^\infty \text{e}^{\ a\ x^2}\ \text dx\ \ \ \ \ ?$$

hold, even if the imaginary part of $b$ is nonzero?

What I really want to understand is what the phrase "By analogy with the previous integrals" means in that link. There, the expression $\frac{J}{a}$ is complex but they seem to imply the integral can be solved like above anyway.

The reusult tells us that the integral is really independend of $J$, which is assumed to be real here. I wonder if we can also generalize this integral to include complex $J$. In case that the shift above is possible, this should work out.

But even if the idea is here to perform that substitution, how to get rid of the complex $a$ to obtain the result. If everything is purely real or imaginary, then this solves the rest of the problem.

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Links to the answer by username RaymondManzoni, see here as well as here. – NikolajK Aug 16 '12 at 18:15
up vote 3 down vote accepted

Let us write $b= r+it$. The real part of $b$ does not matter as you have already proven yourself. So wlog $r=0$.

For shifting along the imaginary axis, we have to employ the residue theorem. We have $$ \begin{align} \int_{-\infty}^\infty f(x+i t) \,dx&- \int_{-\infty}^\infty f(x)\, dx\\ &=\int_{-\infty-it}^{\infty-it} f(x) \,dx- \int_{-\infty}^\infty f(x)\, dx \\ &= 2\pi i \sum \text{Res}(f)+ \int_{\infty-it}^{\infty} f(x) \,dx - \int_{-\infty-it}^{-\infty} f(x) \,dx \end{align},$$ where $\sum \text{Res}(f)$ is the sum over the residues of $f$ in the area $z\in \mathbb{C}$ with $-t<\text{Im}\, z<0$.

So the two integrals are the same if there are no residues and if the two integral at $\pm \infty$ vanish (both of which is the case for your example as long as $\text{Re}\,a <0$).

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Another approach that sometimes simplifies this sort of issue is to invoke the "identity principle": when the integral is holomorphic as a function of the parameter (some potential danger here!), the outcome can be computed in a convenient range, and then invoke the identity principle to know that the same formula holds for all (!) parameter values.

In the case at hand, this approach does succeed.

Beware, in cases like Cauchy's formula for $z$ inside a circle $\gamma$, $f(z)={1\over 2\pi i}\int_\gamma {f(\zeta)\;d\zeta\over \zeta-z}$, the integrand is not holomorphic in the parameter $z$ as it crosses the circle. That is, certainly the integral represents $f(z)$ for $z$ inside, but not outside, where it is $0$.

Also, in the context in which such questions would arise, it might be reasonable to be more careful about "moving contours": note that $\int_{=\infty}^{+\infty}$ is really a limit of integrals $\int_{-M}^N$, and, thus, a contour-shift uses an integral over a rectangle (or parallelogram!) with one side the interval $[-M,N]$. This, too, legitimizes the change of variables for complex parameter values.

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Put $$(x+b)^2=u^2\Longrightarrow 2(x+b)dx=2udu\Longrightarrow dx= du$$ so we get

$$\int_{-\infty}^\infty e^{a(x+b)^2}dx=\int_{-\infty}^\infty e^{au^2}du$$ and your question's answered in the affirmative.

Added: You may want to divide the integral in two rays $\,(-\infty,0)\,,\,(0,\infty)\,$ and then do the above, to avoid problems with the signs after taking the square root.

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You didn't shift the integral path in the complex direction. – NikolajK Aug 16 '12 at 16:10
I'm not sure I need that as the part $\,b\,$ plays in the above is almost as a dummy variable: it disappears altogether when the variable change is done. Yet I must confess I am not sure about what you say. – DonAntonio Aug 16 '12 at 16:13
But if $\Im b\neq 0$, you are moving the integration path (the integral $\int_{-\infty}^\infty$ is to be understood as being on the real line, unless explicitly specified otherwise). This is different to the purely real case where you only move the integration path along the real line (which, given that you integrate along the complete real line, doesn't matter). – celtschk Aug 16 '12 at 16:17
Ok @celtschk , I think I see your point about $\,int_{\infty}^\infty\,$ . I'll take a while to think about this and check whether I can come up with something I consider worthwhile, otherwise I shall delete my answer. Thank you and Nick – DonAntonio Aug 16 '12 at 16:19
@DonAntonio: I think that he is bothered by the real-imaginary rotation in the wikipedia link (the previous paragraph had a real integral now 'by analogy' we got a integral with a $i$ added without changing the path of integration i.e. without multiplying by $i$). This analogy is often explained by analytic continuation... – Raymond Manzoni Aug 16 '12 at 16:19

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