Sine x / x integral: any way to find indefinite integral?

[KrisKringle]

Is there an analytic way to find the indefinite integral of sine(x)/x ?

I've tried several variations of integration by parts, but to no avail.

TIA

Kris

 

[fasteddie65]

Re: Sine x / x integral

No, because it has no antiderivative.
It can only be done graphically.

 

[Deleted member 4993]

Re: Sine x / x integral

Another way would be to express it as a series (Taylor series expansion)

\(\displaystyle \sin(x) \, = \, \sum_{n=0}^{\infty}{{(-1)}^{n}\cdot \frac{x^{(2n+1)}}{n!}}\)

\(\displaystyle \frac{\sin(x)}{x} \, = \, \sum_{n=0}^{\infty}{{(-1)}^{n}\cdot \frac{x^{(2n)}}{n!}}\)

and continue...

 

[KrisKringle]

Re: Sine x / x integral

Thanks to you both (fasteddie65 and subhotosh)

I'd already looked at the series approach, but I couldn't see an obvious way to sum the resulting infinite series.

I strongly suspect that, simple as it looks, this is one that can't be done analytically, and would have to be done numerically.

Thanks for your time and your help.

Regards

Kris

 

[galactus]

Re: Sine x / x integral

If we give it limits, we can do it using differentiation under the integral sign.

\(\displaystyle \int_{-\infty}^{\infty}\frac{sin(x)}{x}dx\)

We can choose a parameter make sure it is Lebesgue integrable(I will skip that part and assume it is...it is).

\(\displaystyle I(a)=\int_{0}^{\infty}\frac{sin(x)}{x}\cdot e^{-ax}dx\)

\(\displaystyle I'(b)=\frac{d}{da}\int_{0}^{\infty}\frac{sin(x)}{x}\cdot e^{-ax}dx\)

\(\displaystyle =\int_{0}^{\infty}\frac{\partial}{{\partial}a}\left[\frac{sin(x)}{x}\cdot e^{-ax}\right]dx\)

\(\displaystyle =\int_{0}^{\infty}sin(x)e^{-ax}dx=\frac{e^{-ax}(cos(x)+a\cdot sin(x))}{1+a^{2}}|_{0}^{\infty}\)

\(\displaystyle =\frac{-1}{1+a^{2}}\)

Integration of I'(a) gives:

\(\displaystyle I(a)=-tan^{-1}(a)+C\)

We choose a strategic value \(\displaystyle a=a_{0}\) in order to make our integrand vanish so that \(\displaystyle I(a_{0})=0\).

In this case, take \(\displaystyle a={\infty}\) so that \(\displaystyle I(\infty)=0 \Rightarrow C=tan^{-1}(\infty)=\frac{\pi}{2}\)

\(\displaystyle \int_{-\infty}^{\infty}\frac{sin(x)}{x}dx=2\int_{0}^{\infty}\frac{sin(x)}{x}dx=2I(0)={\pi}\)

Just a technique I thought you might like.

 

[KrisKringle]

Re: Sine x / x integral

Thanks very much, Galactus. This is really neat.

Kris

 

[daon]

Re: Sine x / x integral

Integrating over (a,b):

\(\displaystyle \lim_{n \rightarrow \infty} \sum _{i=1}^{n}(\frac{b-a}{n}) \cdot \frac{sin(a+\frac{b-a}{n}i)}{(a+\frac{b-a}{n}i)}\)

\(\displaystyle (b-a)\lim_{n \rightarrow \infty}\sum _{i=1}^{n}\frac{sin(a+\frac{b-a}{n}i)}{an+(b-a)i}\)


Some simple VB code that will give you an "almost exact" answer:

        Dim b As Double
        Dim a As Double
        Dim n As Long
        Dim sum As Double
        n = 100001 'use larger n for more precision (choose n odd)
        a = -3.14159 'lower limit   
        b = 3.14159 'upper limit
        sum = 0
        For i = 0 To n ' if one of your limits is a=0 or b=0 start at i=1
            sum = sum + Math.Sin(a + (b - a) / n * i) / (a * n + (b - a) * i)
        Next
        sum=(b-a)*sum
        Call MsgBox( sum)

 

[KrisKringle]

Re: Sine x / x integral

Hi Daon

Thanks for your suggestion. Sorry to get back to you so late, but I basically got the answers I thought I'd get - which was that it can't be done analytically - so I haven't checked this forum for some time. My apologies for not getting back to you sooner.

Best wishes

Kris