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Often in mathematics, particularly in physics, we welcome definite integrals from $-\infty$ to $\infty$ of odd functions, since they are equal to zero. Such as

$\int_{-\infty}^{\infty} \sin(x) dx = 0$ .

So, simple question; why does WolframAlpha fail to evaluate infinite bounded definite integrals of odd functions, stating that the solution "does not converge"? Is it not exactly accurate to say that the answer is zero?

Edit: Rather, I should say that the integral over symmetric bounds, in general, of an odd function is zero. Why when we use bounds at $\pm\infty$ is this not the case?

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I believe Wolfram is throwing an error because $\int_{-\infty}^\infty$ can be interpreted in different ways. You cannot treat $-\infty$ and $\infty$ the same way you treat numbers. Usually in calculus (and physics), infinity shows up in the context of taking some limit as some quantity gets large without bound. The trouble is infinity can be approached in different ways and at different rates ($x^2$ approaches $\infty$ much faster than $x$ for example). And these different ways you can approach infinity are not necessarily equivalent. Thus you get conflicts like "What is $\infty - \infty$?" which can have any number of answers because you can approach one infinity faster than the other.

Because we can approach $\infty$ in different "ways" we must pick a way when we define something like $\int_{-\infty}^\infty$. The standard way is to define it like so: $$ \int_{-\infty}^\infty f(x)\,dx := \lim_{a \to -\infty} \int_a^c f(x)\,dx + \lim_{b \to \infty} \int_c^b f(x)\,dx $$ where $c$ is a constant and can be any real number. You'll notice that if we try to apply this definition for $\sin x$ we run into a problem. For sake of clarity, let me just focus on the first term of the definition (the one with the limit in $a$): \begin{align} \lim_{a \to -\infty} \int_a^c \sin x\,dx &= \lim_{a \to -\infty} \Big[-\cos x \Big]_a^c \\ &= \lim_{a \to -\infty} \big[\cos(a) \big] - \cos(c) \end{align} But this limit doesn't converge because cosine oscillates forever when you approach infinity. Since this limit is a component of the definition I've stated for $\int_{-\infty}^\infty$, it means the expression $\int_{-\infty}^\infty \sin x\,dx$ must be left undefined, and hence Wolfram|Alpha (correctly) barfs.

But of course there is another way you can define $\int_{-\infty}^\infty$. Namely as the Cauchy Principal Value that @Basti mentions. With this definition we define $$ \textrm{p.v.} \int_{-\infty}^\infty f(x)\,dx := \lim_{R \to \infty} \int_{-R}^R f(x)\,dx $$ where the "$\textrm{p.v.}$" is prefixed to distinguish it from the standard definition. Now what happens if we try to use it on $\sin x$? \begin{align} \textrm{p.v.} \int_{-\infty}^\infty \sin x\,dx &= \lim_{R \to \infty} \int_{-R}^R \sin x\,dx \\ &= \lim_{R \to \infty} (0) \quad \textrm{because $\sin$ is an odd function} \\ &= 0 \end{align} which is what you were expecting.

So in short: $\int_{-\infty}^\infty$ can either converge or diverge for some function depending on exactly how you define that expression. Not all sensible definitions are equivalent, so you need to make sure you know which one you're using before you evaluate it. However, as a final note, there are those times when the standard definition and the Cauchy Principal Value do agree. In fact, whenever the standard definition works, the Cauchy Principal Value is guaranteed to converge, and in fact will converge to the same result. The converse is not true in general (as I've shown). Thus the standard definition is a "stronger" notion than the principal value notion: it's harder to satisfy, but it tells you more if it is.

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  • $\begingroup$ Thank you for the great answer. I'm trying to make sure I'm taking away the correct conclusion from this... what exactly do you mean by "Thus the standard definition is a "stronger" notion than the principal value notion: it's harder to satisfy, but it tells you more if it is."? $\endgroup$ – user278703 May 3 '17 at 5:46
  • $\begingroup$ Glad it was helpful :). What I mean is that if the standard definition is satisfied, then you get PV for free, but not vice-versa. In that sense, the standard definition is "stronger" than PV because it automatically includes all the properties of PV plus more. Thus it is "harder" to satisfy than PV because to be standard, you must be everything PV is and more. Often times when an object satisfies a strong notion, you can deduce more about it. E.g. if I told you I was a French president, you could deduce more about me than if I had merely said I am a French citizen. I'm neither, by the way $\endgroup$ – WB-man May 3 '17 at 7:24
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The problem is that $$ \lim_{R\to\infty}\int_{-R}^{R} \sin(x) dx = 0 $$ but $$ \lim_{R,Q\to\infty}\int_{-Q}^{R} \sin(x) dx$$ does not exist. The former is usally called "Cauchy principal value" and is useful in many calculations.

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  • $\begingroup$ Hmmm. I don't see why there is any meaningful difference between the two $\endgroup$ – user278703 May 3 '17 at 3:54
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First try to solve the integral of (e^-nx)sinx from zero to infinity using a suitable reduction formula then put n=0 to get integral of sinx from 0 to infinity =0 then substitute x =-x in the integral and both the integrals to get the desired answer

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