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

TR Smith is a teacher and creator who uses mathematics in her line of work every day.

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The functions f(x) = sin(x)/x and g(x) = cos(x)/x are two of the simplest examples of non-integrable functions in the sense that their antiderivatives cannot be expressed in terms of elementary functions, in other words, they don't have closed-form antiderivatives. 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 function that is non-integrable. (Sometimes these false starts are still useful in discovering other functions that don't have closed-form antiderivatives.)

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 neither sin(x)/x and cos(x)/x have closed-form antiderivatives. Some examples of using approximation formulas to compute integrals are given below, along with some curious definite integrals, and extraction of other functions that do not have closed-form antiderivatives.

## 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.

## Related Integrals

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.

## 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 Functions Using 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

## List of Functions That Cannot Be Exactly Integrated

Here are more examples of non-integrable functions in the sense that they do not have closed-form antiderivatives. They can be found by "massaging" the integrals of sin(x)/x and cos(x)/x with integration techniques.

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 function that cannot be integrated exactly in terms of elementary functions. Likewise, sin(x)*Ln(x) also has no closed-form antiderivative. 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 function with no antiderivative. 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 functions that cannot be integrated in terms of simpler, standard elementary 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 do not have elementary antiderivatives. 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 particularly special functions with names. One reason why they are not so designated is that they do not behave nicely; they are undefined at infinitely many points.

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• sdh 22 months ago

Is there antiderivative of arctan(x)/x?

• Author

TR Smith 22 months ago from Germany

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 22 months ago

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

• Author

TR Smith 22 months ago from Germany

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 10 months ago

That helped me a lot, thank you!

• Aishah 2 months ago

• Clay 7 weeks ago

I believe you that sin(x)/x is non-integrable in terms of trig functions, but I'm having trouble seeing where I go wrong with this chain of substitutions.

x=e^-t → -sin(e^-t)*e^-t dt

t=ln(v) → sin(v)*v

Somehow I started with an non-integrable function and ended up with one that I can integrate with integration by parts?

• Author

TR Smith 7 weeks ago from Germany

Hi Clay, thanks for the question. You have some mistakes in your substitutions. If you start with the integral sin(x)/x dx and make the substitution x =e^-t, you also get dx = -e^-t dt, which gives you the new integral -sin(e^-t) dt.

Then if you do the substitution t = ln(v), you also get dt = 1/v dv, which gives you the new integral -sin(1/v)/v. This does not have a closed-form antiderivative.

If you did one more substitution v =1/y and dv= -1/y^2 dy, you would get the integral sin(y)/y, which is right where you started. Unfortunately, there is no chain of substitutions that will yield something you can integrate in terms of elementary functions.

• Perry 4 weeks ago

Hello!

I was wondering if there's a much more explicit relationship that can be teased out which connects the values of m and n in the following definite integral, evaluated from 0 to infinity...

∫ (sin(x)^m)/x^n dx

...and the integral's actual numerical value.

Thanks so much!

P.