Why the integral of a complex exponential can't be equal to zero?

[kent davidge]

I was just playing with the integral $\int e^{ixa}dx$ when I found something interesting. If you integrate from $x = m2\pi/a$ to $x = n2\pi/a$ where $m$ and $n$ are any two integers, the integral equals zero.

On one hand, as we can in principle choose whatever values we like for $m$ and $n$ (as long as they are integer numbers), if we let $m \rightarrow -\infty$ and $n \rightarrow +\infty$, then the integral $\int_{-\infty}^{+\infty} e^{ixa}dx$ should vanish.

On the other hand, this is absurd, since I know these exponentials even work as an orthonormal basis in Fourier expansions.

I presume my mistake is in taking those limits. What you think?

 

[BvU]

if we let $m \rightarrow -\infty$ and $n \rightarrow +\infty$, then the integral $\int_{-\infty}^{+\infty} e^{ixa}dx$ should vanish.

No it won't. Why should it ?

this is absurd, since I know these exponentials even work as an orthonormal basis

What's the contradiction ?

 

[kent davidge]

Why should it ?

because the limits of integration would be like $-\infty$ and $+\infty$

What's the contradiction ?

if the mentioned integral vanished, we couldn't use the exponentials as bases at all.

 

[BvU]

because the limits of integration would be like $-\infty$ and $+\infty$

That is not a criterion. The integrand does not go to zero.

 

[Infrared]

You're not allowed to restrict $x$ to be of the form $k\pi/a$ when taking a limit $x\to\infty$. For a simpler example, it's not true that $\lim_{n\to\infty}(-1)^n=1$ even though $(-1)^n=1$ when $n$ is even. Also note that you're assuming $a\neq 0$.

That is not a criterion. The integrand does not go to zero.

Integrable functions don't have to vanish at infinity. Consider $f(x)=1$ when $n\leq x\leq n+2^{-n}$ for natural $n$, and $0$ elsewhere. You can easily modify this to make a continuous counterexample if you'd like.

 

[FactChecker]

because the limits of integration would be like $-\infty$ and $+\infty$

You are "cherry-picking" particular values for the range of the integral to get 0. That is not the proper way to take the limit of the integral as the upper and lower bounds go to infinity.

 

[mathwonk]

apparently you are taking an integral over the real axis, so your integral vanishes if and only if the real and imaginary parts do. so graph say the real part of your integral, essentially cos(x), and look at the behavior as x goes to plus or minus infinity. You see the total amount of signed area oscillates back and forth, and equals zero infinitely often, but has no limit at infinity.