Evaluation of improper integral involving sinx/x

In summary: Therefore, the integral with the absolute value sign is infinite. In summary, the integral of |sinx/x| from 0 to infinity is infinite when evaluated with the absolute value sign, but is a Dirichlet integral when evaluated without it. The proof for its infinity is shown through comparison with a harmonic series.
  • #1
ashok vardhan
19
0
Sir,

Recently when i am evaluating a convolution integral, i came across the integral of |sinx/x| under limits running from 0 to infinity. when i tried to evaluate the integral, i used complex analysis tools like assuming a function e^(iz) / z and deduce the above integral from integral
of f(z) by contour integration.But, i am unable to arrive at any conclusion.I wanted to know whether the integral is finite or not.Can you please help me.
 
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  • #2
Do you mean the integral with or without the absolute value sign in the integrand? You use the absolute value in one place but not the other.

Without the absolute value, this is a Dirichlet integral. Evaluating it with a few different methods (including contour integration) is shown in the link.
 
  • #3
With the absolute value, the integral is infinite. To see this, observe that
$$\left|\frac{\sin(x)}{x}\right| > \left|\frac{1}{2x}\right|$$
if and only if
$$\left|\sin(x)\right| > \frac{1}{2}$$
if and only if
$$x \in \left[\frac{\pi}{6}, \frac{5\pi}{6}\right] + n\pi$$
where ##n## is any integer. Therefore,
$$\int_{0}^{\infty}\left|\frac{\sin(x)}{x}\right| dx \geq \sum_{n=0}^{\infty}\int_{\frac{\pi}{6} + n\pi}^{\frac{5\pi}{6}+n\pi} \left|\frac{1}{2x}\right| dx \geq \sum_{n=0}^{\infty} \frac{2\pi}{3} \left(\frac{1}{\frac{\pi}{3} + 2n\pi}\right)$$
and the rightmost expression is infinite by comparison with a harmonic series.
 

Related to Evaluation of improper integral involving sinx/x

1. What is an improper integral?

An improper integral is an integral that does not have a finite value due to either the limits of integration being infinite, or the integrand having an infinite discontinuity within the interval of integration.

2. How do you determine if an improper integral involving sinx/x is convergent or divergent?

An improper integral involving sinx/x is convergent if the limit of the integral as the upper limit approaches infinity is finite. It is divergent if the limit is either infinite or does not exist.

3. Can an improper integral involving sinx/x be evaluated using the fundamental theorem of calculus?

No, the fundamental theorem of calculus can only be applied to integrals with finite values. Improper integrals require special techniques to evaluate them.

4. What are some common techniques for evaluating improper integrals involving sinx/x?

Some common techniques include using limits of integration to split the integral into two or more finite integrals, using trigonometric identities to simplify the integrand, and using integration by parts.

5. Can an improper integral involving sinx/x have a different value at different limits of integration?

Yes, an improper integral can have a different value depending on the limits of integration. This is because the value of the integral is affected by the behavior of the integrand at the limits, and the limit of the integral may exist at one limit but not at another.

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