Tag: Parameter Integrals

A Surprising Improper Integral

Consider the improper integral

$${\int }_0^{\infty }\frac{\sin (ax)}{x}dx.$$

At first glance, this integral looks difficult. The factor $\frac{1}{x}$ suggests a singularity at the origin, while the sine function continues to oscillate forever as $x\to \infty$. There is no elementary antiderivative that immediately resolves the problem.

Nevertheless, for every positive number $a$, the answer is remarkably simple:

$${\int }_0^{\infty }\frac{\sin (ax)}{x}dx=\frac{\pi }{2}.$$

Even more surprisingly, the answer does not depend on the positive value of $a$. Let us see why.

The Main Idea: Add a Damping Factor

Instead of attacking the original integral directly, introduce a positive parameter $t$ and define

$$F(t)={\int }_0^{\infty }{e}^{-tx}\frac{\sin (ax)}{x}dx,t>0.$$

The factor $e^{-tx}$ suppresses the oscillations for large values of $x$. This makes the parameter-dependent integral easier to work with.

The key step is to differentiate with respect to the parameter $t$. We obtain

$$F^{\prime }(t)=-{\int }_0^{\infty }{e}^{-tx}\sin (ax)dx.$$

Notice what happened: the troublesome factor $\frac{1}{x}$ has disappeared. We are left with a standard Laplace-type integral.

Evaluating the Easier Integral

For $t>0$, we have

$${\int }_0^{\infty }{e}^{-tx}\sin (ax)dx=\frac{a}{t^2+a^2}.$$

Therefore,

$$F^{\prime }(t)=-\frac{a}{t^2+a^2}.$$

Now the problem has been reduced to an elementary integral.

Recovering F(t)

As $t\to \infty$, the exponential damping becomes stronger and

$$F(t)\to 0.$$

Thus, for $a>0$,

$$F(t)={\int }_t^{\infty }\frac{a}{u^2+a^2}du.$$

Evaluating this integral gives

$$F(t)=\frac{\pi }{2}-\arctan \left(\frac{t}{a}\right).$$

Equivalently, using the elementary arctangent identity,

$$F(t)=\arctan \left(\frac{a}{t}\right).$$

So we have actually obtained the more general and useful formula

$${\int }_0^{\infty }{e}^{-tx}\frac{\sin (ax)}{x}dx=\arctan \left(\frac{a}{t}\right),a>0,t>0.$$

Removing the Damping

We introduced the exponential factor only to make the integral easier to evaluate. Now let $t\to 0^{+}$. Then

$$\arctan \left(\frac{a}{t}\right)\to \frac{\pi }{2}.$$

and the damping factor approaches $1$. This leads to the celebrated Dirichlet integral

$${\int }_0^{\infty }\frac{\sin (ax)}{x}dx=\frac{\pi }{2},a>0.$$

Why Does the Answer Not Depend on a?

There is also a simple scaling argument that explains why the answer must be the same for every positive $a$. Set

$$u=ax.$$

Then

$$x=\frac{u}{a},dx=\frac{du}{a}.$$

Therefore,

$${\int }_0^{\infty }\frac{\sin (ax)}{x}dx={\int }_0^{\infty }\frac{\sin (u)}{u}du.$$

The parameter $a$ has completely disappeared. Changing $a$ changes how rapidly the sine function oscillates, but the total value of the improper integral remains unchanged.

What If a Is Zero or Negative?

If $a=0$, the integrand is identically zero, so the integral is zero.

If $a<0$, use the oddness of the sine function:

$$\sin (ax)=-\sin (-ax).$$

Hence the complete result is

$${\int }_0^{\infty }\frac{\sin (ax)}{x}dx=\left\{\begin{array}{ll}\frac{\pi }{2}& \text{if}a>0,\\ 0& \text{if}a=0,\\ -\frac{\pi }{2}& \text{if}a<0.\end{array}\right.$$

One Important Detail: The Integral Is Not Absolutely Convergent

The convergence of this integral is subtle. Although

$${\int }_0^{\infty }\frac{\sin (ax)}{x}dx$$

converges for $a\ne 0$, the corresponding absolute-value integral

$${\int }_0^{\infty }\frac{|\sin (ax)|}{x}dx$$

diverges. Thus the positive and negative oscillations of the sine function are essential. They cancel one another just enough for the original improper integral to converge.

A Useful Lesson

The most interesting part of this calculation is not simply the final answer $\frac{\pi }{2}$. It is the method.

When an integral is difficult to evaluate directly, it can sometimes be embedded into a family of integrals depending on a parameter. Differentiating with respect to that parameter may transform the original problem into a much easier one. After solving the parameterized problem, we return to the original integral by taking a limit.

In this example, the chain of ideas is

$$\frac{\sin (ax)}{x}\to e^{-tx}\frac{\sin (ax)}{x}\to F^{\prime }(t)\to F(t)\to \frac{\pi }{2}.$$

A difficult oscillatory improper integral has been reduced to an elementary rational integral. That is what makes the Dirichlet integral such a beautiful example of the power of introducing a parameter.

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Another Surprise from an Improper Integral

The integral involving $\frac{\sin \left(ax\right)}{x}$ shows that an oscillating function extending over an infinite interval can nevertheless produce a beautifully simple finite value.

There is another famous improper-integral paradox in which infinity appears in a completely different way: a surface extending forever can enclose a finite volume while having infinite surface area.

Continue exploring: Gabriel’s Horn: When Can an Infinite Horn Be Painted?

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From the Dirichlet Integral to the Gaussian Integral

There is another beautiful connection behind the Dirichlet integral. The Gaussian function $e^{-x^2}$ leads to one of the most famous improper integrals in mathematics. Its evaluation introduces a remarkably powerful idea: turn a one-dimensional integral into a two-dimensional one and then use geometry.

That same Gaussian structure appears in many unexpected places and provides another route into the world of remarkable improper integrals.

Continue exploring: The Gaussian Integral and Beyond: From e^(-x²) to a Family of Integrals