How to Integrate sin(x)/x and cos(x)/x

The functions f(x) = sin(x)/x and g(x) = cos(x)/x are two of the simplest examples of non-integrable functions, meaning their antiderivatives cannot be expressed in terms of elementary functions. No matter how you attempt u-substitution or integration by parts on sin(x)/x or cos(x)/x, you will always end up with another non-integrable function. (Sometimes these false starts are still useful in discovering other non-integrable functions.)

Nevertheless, if you want to compute a definite integral of sin(x)/x or cos(x)/x, you can use approximation formulas based on a truncation of their infinite power series expansions.

Related to sin(x)/x and cos(x)/x are the functions Si(x) and Ci(x), known as the Sine Integral and Cosine Integral functions respectively. They are defined as

Si(x) = ∫ sin(t)/t dt, from t = 0 to t = x
Ci(x) = -∫ cos(t)/t dt, from t = x to t = infinity

Neither Si(x) nor Ci(x) has a closed-form representation since sin(x)/x and cos(x)/x are non-integrable. Some examples of using approximation formulas to compute integrals are given below, along with some curious definite integrals, and extraction of other non-integrable trig functions.

Approximate Antiderivative of sin(x)/x

The Taylor series expansion of sin(x) is

sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...

If we divide this by x, we can obtain a power series expansion for sin(x)/x:

sin(x)/x = 1 - (x^2)/3! + (x^4)/5! - (x^6)/7!+ ...

Integrating each term of the series gives you an infinite series for the integral of sin(x)/x:

∫ sin(x)/x dx = x - (x^3)/(3*3!) + (x^5)/(5*!) - (x^7)/(7*7!) + ... + C

If we truncate this series after at least four terms we can obtain a pretty good approximation formula for the antiderivative of sin(x)/x. The more terms are included, the more accurate the approximation will be, since this is a convergent power series. We can now use this formula to estimate the area under the curve y = sin(x)/x in the following example.

Example 1: Area Under y = sin(x)/x

The graph above shows the curve y = sin(x)/x with the area from x = 0 to x = π shaded in yellow. What is the approximate area of this region? To solve this calculus problem we use the integral approximation formula that comes from truncating the power series after the fifth term. This gives us

∫ sin(x)/x dx ≈ x - (x^3)/18 + (x^5)/600 - (x^7)/35280 + (x^9)/3265920 + C

Plugging x = 0 and x = π into this expression and subtracting gives us

π - (π^3)/18 + (π^5)/600 - (π^7)/35280 + (π^9)/3265920 ≈ 1.852.

Approximate Antiderivative of cos(x)/x

The Taylor series expansion of cos(x) is

cos(x) = 1 - (x^2)/2! + (x^4)/4! - (x^6)/6! + ...

Dividing this by x gives us an infinite series expansion for cos(x)/x:

cos(x)/x = 1/x - x/2! + (x^3)/4! - (x^5)/6! + ...

And finally, integrating this series term by term gives us a power series expansion for the integral of cos(x)/x:

∫ cos(x)/x dx = Ln(x) - (x^2)/(2*2!) + (x^4)/(4*4!) - (x^6)/(6*6!) + ... + C

This differs from the series expansion of the antiderivative of sin(x)/x because we have a natural logarithm as one of the terms. Nevertheless, we can also use this series to estimate the definite integral of cos(x)/x.

Example 2: Area Under y = cos(x)/x

The graph above shows the curve y = cos(x)/x with the area between x = π/2 and x = 3π/2 shaded in yellow. We will use the infinite series for the antiderivative to estimate the area of this region. To do this, we plug the values of x = 3π/2 and x = π/2 into the truncated series

Ln(x) - (x^2)/4 + (x^4)/96 - (x^6)/4320 + (x^8)/322560

and subtract the two values. Doing this gives us

[ Ln(3π/2) - ((3π/2)^2)/4 + ((3π/2)^4)/96 - ((3π/2)^6)/4320 + ((3π/2)^8)/322560 ]
- [ Ln(π/2) - ((π/2)^2)/4 + ((π/2)^4)/96 - ((π/2)^6)/4320 + ((π/2)^8)/322560 ]

≈ 0.540.

Summary of Approximate Antiderivatives

Contour integral path for f(x) = sin(x)/x.
Contour integral path for f(x) = sin(x)/x.

A Convergent Integral

The improper definite integral of sin(x)/x from x = 0 to x = infinity converges and is equal to π/2. This is not a result that can be derived with standard single-variable integral calculus techniques. Instead it is found using contour integration, a technique in complex-variable calculus. This result can be extended to find other amazingly elegant definite integral forms, several of which are listed below.

Definite improper integrals of the form ∫ (sin(x)^m)/x^n dx, evaluated from x = 0 to x = infinity, will converge so long as m is greater than or equal to n. If n = 2 and m is even, the integral ∫ (sin(x)^m)/x^n dx {x = 0 to x = ∞} will converge to a rational multiple of π. If n = 2 and m is odd, the integral will converge to an expression involving natural logarithms of integers.

Finding Other Non-Integrable Functions With sin(x)/x & cos(x)/x

Suppose we attempt to reduce the integral of sin(x)/x using integration by parts, letting u = sin(x) and dv = 1/x dx. This gives us du = cos(x) dx and v = Ln(x). Plugging these pieces into the integration by parts formula gives us

∫ sin(x)/x = sin(x)*Ln(x) - ∫ cos(x)*Ln(x) dx

Non-Integrable Function List

These are a few examples of non-integrable functions that can be found by massaging the integrals of sin(x)/x and cos(x)/x.

f(x) = sin(x)*Ln(x)
f(x) = cos(x)*Ln(x)
f(x) = sin(1/x) / x
f(x) = cos(1/x) / x
f(x) = sin(x^2)/x
f(x) = cos(x^2)/x
f(x) = sin(x^m)/x^n, positive m and n
f(x) = cos(x^m)/x^n, positive m and n
f(x) = x/[sqrt(1-x^2)*arcsin(x)]
f(x) = x/[sqrt(1-x^2)*arccos(x)]
f(x) = x*sin(x)*Ln(x)
f(x) = x*cos(x)*Ln(x)

Since the function we started with can't be integrated in terms of elementary functions, neither can cos(x)*Ln(x); thus, we have discovered another non-integrable function. Likewise, sin(x)*Ln(x) is also non-integrable. Let's try integration by parts again, but this time using the assignment u = 1/x and dv = sin(x) dx, which also gives us du = -1/x^2 and v = -cos(x). This yields

∫ sin(x)/x = -cos(x)/x - ∫ cos(x)/x^2 dx

And so we can see that cos(x)/x^2 has no antiderivative, and likewise for sin(x)/x^2.

Now suppose we attempt u-substitution on the integral of sin(x)/x using x = 1/u and dx = -1/u^2 du. This gives us

∫ sin(x)/x = -∫ sin(1/u)/u du

So we find that sin(1/x)/x is another non-integrable function, as is cos(1/x)/x. Let's try the substitution x = arcsin(u) and dx = 1/sqrt(1 - u^2) du. This yields

∫ sin(x)/x = ∫ u/[sqrt(1 - u^2)*arcsin(u)] du

which is yet another non-integrable function. The upshot of all this seemingly dead-end work is that even though standard integration techniques proved futile in finding the antiderivative of sin(x)/x and cos(x)/x, they proved useful in discovering more non-integrable functions.

What About the Integrals of tan(x)/x and sec(x)/x?

Like the integrals of sin(x)/x and cos(x)/x, the integrals of tan(x)/x and sec(x)/x cannot be expressed in terms of elementary functions. But unlike sin(x)/x and cos(x)/x, whose integrals are designated as the special functions Si(x) and Ci(x), the antiderivatives of tan(x)/x and sec(x)/x are not considered to be any particularly special functions. One reason why they are not so designated is that they do not 'behave nicely' as functions and are undefined at infinitely many points.

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Comments 6 comments

sdh 20 months ago

Is there antiderivative of arctan(x)/x?

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calculus-geometry 20 months ago from Germany Author

The antiderivative of arctan(x)/x cannot be expressed in terms of elementary functions, but the antiderivative of arctan(x)/x^n does have a closed-form expression for integers n greater than 1.

Noris 19 months ago

Which functions of the form Sin[X^m]X^n have antiderivatives?

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calculus-geometry 19 months ago from Germany Author

Here are some cases where you can find the antiderivative of sin(x^m)*x^n:

n = m-1

m = 1, and n = non-negative integer

m = 1/p and n = q/p, where p and q are positive integers

There are other cases you can find by applying u-substitution.

freethinker07 7 months ago

That helped me a lot, thank you!

Aishah 5 days ago

Very helpful thank you

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