# Integration of sinx/x

1. Sep 24, 2011

### gatorules

Wherever I look on the internet it says that sinx/x cannot be integrated by elementary techniques, but it seems that there is a method using integration by parts of the quotient rule. However, when I compute definite integrals with this, the answer that my calculator returns is different than the answer I get with the definite integral. Can anyone tell me where my problem is or if my method does actually work.

$$\frac{d}{dx}\frac{f(x)}{g(x)} = \frac{f'(x)g(x)-g'(x)f(x)}{g^2(x)}$$
$$\frac{f(x)}{g(x)}=\int \frac{f'(x)g(x)}{g^2(x)}\,dx-\int \frac{g'(x)f(x)}{g^2(x)}\,dx$$
$$\int \frac{g'(x)f(x)}{g^2(x)}\,dx=\int \frac{f'(x)}{g(x)}\,dx-\frac{f(x)}{g(x)}\,dx$$
$$\int \frac{sin(x)}{x}\,dx=\int \frac{xsin(x)}{x^2}\,dx=\frac{1}{2}\int \frac{2xsin(x)}{x^2}\,dx$$
$$g(x)=x^2$$
$$f(x)=sin(x)$$
$$g'(x)=2x$$
$$f'(x)=cos(x)$$
$$\int \frac{g'(x)f(x)}{g^2(x)}\,dx=\int \frac{f'(x)}{g(x)}\,dx-\frac{f(x)}{g(x)}\,dx$$
$$\frac{1}{2}\int \frac{2xsin(x)}{x^2}\,dx=\frac{1}{2}\int \frac{cos(x)}{x}\,dx-\frac{sin(x)}{x}\,dx$$
$$\int \frac{cos(x)}{x}\,dx=\frac{1}{2}\int \frac{2xcos(x)}{x^2}\,dx$$
$$\frac{1}{2}\int \frac{2xcos(x)}{x^2}\,dx=\frac{1}{2}\int \frac{-sin(x)}{x}\,dx-\frac{1}{2}\frac{cos(x)}{x}\,dx$$
$$\frac{1}{2}\int \frac{2xcos(x)}{x^2}\,dx=\frac{1}{2}\int \frac{-sin(x)}{x}\,dx-\frac{1}{2}\frac{cos(x)}{x}\,dx$$
$$\int \frac{sin(x)}{x}\,dx=\frac{1}{2}\left(\frac{1}{2} \int \frac{-sin(x)}{x} \,dx - \frac{1}{2}\frac{cos(x)}{x}\right)-\frac{1}{2}\frac{sin(x)}{x}$$
$$\frac{5}{4}\int \frac{sin(x)}{x}\,dx=-\frac{1}{4}\frac{cos(x)}{x}-\frac{1}{2}\frac{sin(x)}{x}$$
$$\int \frac{sin(x)}{x}\,dx=-\frac{1}{5}\frac{cos(x)}{x}-\frac{2}{5}\frac{sin(x)}{x}$$

2. Sep 24, 2011

### rude man

I got a bit of a headache following you since there are extra dx's everywhere & at least one is missing ....

But, could you check your result against a power series expansion of
sin(x)/x = 1 - x^2/3! + x^4/5! - ....

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

and compare that with the series for your answer?

3. Sep 24, 2011

### gb7nash

I haven't looked at your method yet, but there's an easy way to tell if your method works. Take the derivative of your answer and see if you get sin(x)/x

4. Sep 24, 2011

### rude man

good point gb nash! Somebody please do it - I gotta go hike.

5. Sep 24, 2011

### Dickfore

your mistake is that you took $g(x) = x^{2}$, whereas, according to the formula you derived the denominator is $g^{2}(x)$ and should be equal to $x^{2}$, thus $g(x) = x$. But, then, $g'(x) = 1$ and not $2 x$ as you had assumed.

6. Sep 24, 2011

### gatorules

Thanks. I guess that means the method is totally trashed by that explains why it wasn't working.

7. Sep 24, 2011

### Dickfore

Yes.

Alternatively, you could have used the ordinary integration by parts:

$$\int{\frac{\sin{x}}{x} \, dx} = -\int{\frac{1}{x} \, d(\cos{x})} = -\frac{\cos{x}}{x} - \int{\frac{\cos{x}}{x^{2}} \, dx} = -\frac{\cos{x}}{x} - \int{\frac{1}{x^{2}} \, d(\sin{x})} = -\frac{\cos{x}}{x} - \frac{\sin{x}}{x^{2}} - 2 \int{\frac{\sin{x}}{x^{3}} \, dx}$$

You may notice a recursion for the integrals:
$$I_{n} \equiv \int{\frac{\sin{x}}{x^{2 n + 1}} \, dx}$$

By similar double integration by parts, you get:
$$I_{n} = -\frac{\cos{x}}{x^{2 n + 1}} - \frac{(2 n + 1) \sin{x}}{x^{2 n + 2}} - (2 n + 1) (2 n + 2) I_{n + 1}$$

As you can see, this procedure goes on indefinitely and you do not get a closed form.

8. Sep 25, 2011

### gatorules

Ohhh. That's genius. Thanks.