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# integration of sinx/x with respect to x

bhanuveer danduboyina
95 Points
9 years ago

The fact of the matter is, this integral is one that cannot be expressed in terms of elementary functions. There's no way we can solve this using the methods we know; we cannot use integration by parts, partial fractions, substitution, trigonometric substitution, etc to solve this.

We can, however, approximate the integral through a power series. sin(x) has its own power series, so all we need to do is divide each term of the series by x (this represents (1/x)sin(x), or sin(x)/x) and then integrate thereafter.

como: Your proposed solution doesn't work, and here's why. Let

f(x) = (1/x)cos x - (1/x²)sinx + C

To make it easier to differentiate, factor (1/x).

f(x) = (1/x) [cos(x) - (1/x)sin(x)] + C

Differentiate using the product rule, noting that d/dx (1/x) = -1/x^2 gives us

f'(x) = (-1/x^2) [cos(x) - (1/x)sin(x)] + (1/x) [-sin(x) - [(-1/x^2)sin(x) + (1/x)cos(x)] ]

f'(x) = -cos(x)/x^2 + sin(x)/x^3 - sin(x)/x + sin(x)/x^2 - cos(x)/x

And as you can see, it looks nothing like sin(x)/x.

Rakib
15 Points
one year ago
Alternatively, you could have used the ordinary integration by parts: ∫ sin x x d x = − ∫ 1 x d ( cos x ) = − cos x x − ∫ cos x x 2 d x = − cos x x − ∫ 1 x 2 d ( sin x ) = − cos x x − sin x x 2 − 2 ∫ sin x x 3 d x ∫sin⁡xxdx=−∫1xd(cos⁡x)=−cos⁡xx−∫cos⁡xx2dx=−cos⁡xx−∫1x2d(sin⁡x)=−cos⁡xx−sin⁡xx2−2∫sin⁡xx3dx You may notice a recursion for the integrals: I n ≡ ∫ sin x x 2 n + 1 d x In≡∫sin⁡xx2n+1dx By similar double integration by parts, you get: I n = − cos x x 2 n + 1 − ( 2 n + 1 ) sin x x 2 n + 2 − ( 2 n + 1 ) ( 2 n + 2 ) I n + 1 In=−cos⁡xx2n+1−(2n+1)sin⁡xx2n+2−(2n+1)(2n+2)In+1 As you can see, this procedure goes on indefinitely and you do not get a closed form.