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Daniele Tampieri
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I have recently run into a number of divergent oscillating integrals in various contexts. Thus, I have been led to desire general methods for assigning values to divergent oscillating integrals. All of the integrals I am interested in have the following form $$ \int_0^\infty f(x) \sin(x) dx \text{ or } \int_0^\infty f(x) \cos(x) dx $$ Where $f(x)$ is usually an eventually monotonically increasing function.

Background

In the case where $f(x)$ grows like a polynomial I believe there are various approaches that all provide the same value. For instance, consider
$$\int_0^\infty \sin(x) dx= \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x)\right) dx = \frac{1}{2}\int_0^{\pi} \sin(x)dx + \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x + \pi)\right) dx = 1$$$$ \begin{split} \int_0^\infty \sin(x) dx &= \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x)\right) dx \\ &= \frac{1}{2}\int_0^{\pi} \sin(x)dx + \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x + \pi)\right) dx = 1. \end{split}$$ Alternatively, in an analgous manner to applying a smooth cutoff function to a divergent series, we can also apply a smooth cutoff function to get $$ \lim_{\varepsilon\to 0} \int_0^\infty \sin(x) (1-\varepsilon)^{1+x} dx = 1$$ An analogue to Cesaro summation for integrals also provides the same values for this integral.

We can apply these method to higher powers of $x$ as well. Doing this generates the following values $$ \int_0^\infty x^n \sin(x) = \cos\left( \frac{n \pi}{2}\right) \Gamma(n+1), \int_0^\infty x^n \cos(x) = -\sin\left( \frac{n \pi}{2}\right) \Gamma(n+1)$$

Thus, if $f(x)$ has a power series presentation, we can write $$\int_0^\infty f(x) \sin(x) = \int_0^\infty \sum_{n=0}^\infty \frac{f^{n}(0)}{n!} x^n \sin(x) = \sum_{n=0}^\infty (-1)^n f^{2n}(0)$$ This formula is a good start, but the series typically doesn't converge.

Is there a general way to assign a value to divergent oscillating integrals, especially those where $f(x)$ grows at an exponential rate or faster? Since it seems unlikely that there might be a way to assign a value to a general function-- are there intereting/broad categories of functions which can be assigned values?

I have recently run into a number of divergent oscillating integrals in various contexts. Thus, I have been led to desire general methods for assigning values to divergent oscillating integrals. All of the integrals I am interested in have the following form $$ \int_0^\infty f(x) \sin(x) dx \text{ or } \int_0^\infty f(x) \cos(x) dx $$ Where $f(x)$ is usually an eventually monotonically increasing function.

Background

In the case where $f(x)$ grows like a polynomial I believe there are various approaches that all provide the same value. For instance, consider
$$\int_0^\infty \sin(x) dx= \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x)\right) dx = \frac{1}{2}\int_0^{\pi} \sin(x)dx + \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x + \pi)\right) dx = 1$$ Alternatively, in an analgous manner to applying a smooth cutoff function to a divergent series, we can also apply a smooth cutoff function to get $$ \lim_{\varepsilon\to 0} \int_0^\infty \sin(x) (1-\varepsilon)^{1+x} dx = 1$$ An analogue to Cesaro summation for integrals also provides the same values for this integral.

We can apply these method to higher powers of $x$ as well. Doing this generates the following values $$ \int_0^\infty x^n \sin(x) = \cos\left( \frac{n \pi}{2}\right) \Gamma(n+1), \int_0^\infty x^n \cos(x) = -\sin\left( \frac{n \pi}{2}\right) \Gamma(n+1)$$

Thus, if $f(x)$ has a power series presentation, we can write $$\int_0^\infty f(x) \sin(x) = \int_0^\infty \sum_{n=0}^\infty \frac{f^{n}(0)}{n!} x^n \sin(x) = \sum_{n=0}^\infty (-1)^n f^{2n}(0)$$ This formula is a good start, but the series typically doesn't converge.

Is there a general way to assign a value to divergent oscillating integrals, especially those where $f(x)$ grows at an exponential rate or faster? Since it seems unlikely that there might be a way to assign a value to a general function-- are there intereting/broad categories of functions which can be assigned values?

I have recently run into a number of divergent oscillating integrals in various contexts. Thus, I have been led to desire general methods for assigning values to divergent oscillating integrals. All of the integrals I am interested in have the following form $$ \int_0^\infty f(x) \sin(x) dx \text{ or } \int_0^\infty f(x) \cos(x) dx $$ Where $f(x)$ is usually an eventually monotonically increasing function.

Background

In the case where $f(x)$ grows like a polynomial I believe there are various approaches that all provide the same value. For instance, consider
$$ \begin{split} \int_0^\infty \sin(x) dx &= \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x)\right) dx \\ &= \frac{1}{2}\int_0^{\pi} \sin(x)dx + \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x + \pi)\right) dx = 1. \end{split}$$ Alternatively, in an analgous manner to applying a smooth cutoff function to a divergent series, we can also apply a smooth cutoff function to get $$ \lim_{\varepsilon\to 0} \int_0^\infty \sin(x) (1-\varepsilon)^{1+x} dx = 1$$ An analogue to Cesaro summation for integrals also provides the same values for this integral.

We can apply these method to higher powers of $x$ as well. Doing this generates the following values $$ \int_0^\infty x^n \sin(x) = \cos\left( \frac{n \pi}{2}\right) \Gamma(n+1), \int_0^\infty x^n \cos(x) = -\sin\left( \frac{n \pi}{2}\right) \Gamma(n+1)$$

Thus, if $f(x)$ has a power series presentation, we can write $$\int_0^\infty f(x) \sin(x) = \int_0^\infty \sum_{n=0}^\infty \frac{f^{n}(0)}{n!} x^n \sin(x) = \sum_{n=0}^\infty (-1)^n f^{2n}(0)$$ This formula is a good start, but the series typically doesn't converge.

Is there a general way to assign a value to divergent oscillating integrals, especially those where $f(x)$ grows at an exponential rate or faster? Since it seems unlikely that there might be a way to assign a value to a general function-- are there intereting/broad categories of functions which can be assigned values?

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Caleb Briggs
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Assigning values to a divergent oscillating integrals

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Caleb Briggs
  • 1.7k
  • 1
  • 8
  • 25

Assigning values to a divergent oscillating integrals

I have recently run into a number of divergent oscillating integrals in various contexts. Thus, I have been led to desire general methods for assigning values to divergent oscillating integrals. All of the integrals I am interested in have the following form $$ \int_0^\infty f(x) \sin(x) dx \text{ or } \int_0^\infty f(x) \cos(x) dx $$ Where $f(x)$ is usually an eventually monotonically increasing function.

Background

In the case where $f(x)$ grows like a polynomial I believe there are various approaches that all provide the same value. For instance, consider
$$\int_0^\infty \sin(x) dx= \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x)\right) dx = \frac{1}{2}\int_0^{\pi} \sin(x)dx + \frac{1}{2}\int_0^\infty \left(\sin(x) + \sin(x + \pi)\right) dx = 1$$ Alternatively, in an analgous manner to applying a smooth cutoff function to a divergent series, we can also apply a smooth cutoff function to get $$ \lim_{\varepsilon\to 0} \int_0^\infty \sin(x) (1-\varepsilon)^{1+x} dx = 1$$ An analogue to Cesaro summation for integrals also provides the same values for this integral.

We can apply these method to higher powers of $x$ as well. Doing this generates the following values $$ \int_0^\infty x^n \sin(x) = \cos\left( \frac{n \pi}{2}\right) \Gamma(n+1), \int_0^\infty x^n \cos(x) = -\sin\left( \frac{n \pi}{2}\right) \Gamma(n+1)$$

Thus, if $f(x)$ has a power series presentation, we can write $$\int_0^\infty f(x) \sin(x) = \int_0^\infty \sum_{n=0}^\infty \frac{f^{n}(0)}{n!} x^n \sin(x) = \sum_{n=0}^\infty (-1)^n f^{2n}(0)$$ This formula is a good start, but the series typically doesn't converge.

Is there a general way to assign a value to divergent oscillating integrals, especially those where $f(x)$ grows at an exponential rate or faster? Since it seems unlikely that there might be a way to assign a value to a general function-- are there intereting/broad categories of functions which can be assigned values?