What's the Antiderivative of $\tan(x)/x$?

[eddybob123]

Hi, I am just wondering what the antiderivative of this integral was. It looks easy to me, but I no matter what I did just could not evaluate it. Can someone help me?:
$$\int \frac{\tan(x)}{x}\;dx$$

 

[MarneMath]

Have you tried integration by parts? It can be a bit tedious but it isn't to bad.

 

[TheFool]

Actually, that integral isn't so simple. It cannot be stated as a finite combination of elementary functions.

 

[Goa'uld]

In fact, that is true for all trig functions over x.

 

[eddybob123]

How about using the Fourier series for tan?

 

[Goa'uld]

It depends on the application. If this is part of a larger problem it may be more convenient to define a new function in terms of the integral rather than write an infinite series repeatedly.
$$f'(x)=\frac{\tan{x}}{x}$$
$$f(x)=\int_{0}^{x}\frac{\tan{t}}{t}dt$$
One problem that may occur with infinite series is the radius of convergence. While tan(x)/x may have a value at certain values of x, the infinite series may not converge.

 

[babysnatcher]

Hi, I am just wondering what the antiderivative of this integral was. It looks easy to me, but I no matter what I did just could not evaluate it. Can someone help me?:
$$\int \frac{\tan(x)}{x}\;dx$$

Use integration by parts twice.

 

[Goa'uld]

Use integration by parts twice.

Please demonstrate.

 

[MarneMath]

Bleh, let dv = 1/x u = tanx for the first integration by parts.

Then with the second integral you get let dv = 1/x and dv = sec^2(x)dx. You'll notice that after you do integration by parts twice, that you'll have the same integral on both sides, so combine like terms and come out with ln(x)tan(x)

 

[Goa'uld]

Bleh, let dv = 1/x u = tanx for the first integration by parts.

Then with the second integral you get let dv = 1/x and dv = sec^2(x)dx. You'll notice that after you do integration by parts twice, that you'll have the same integral on both sides, so combine like terms and come out with ln(x)tan(x)

You are mistaken.
$$\frac{d}{dx}(\log{x}\tan{x})=\log{x}\sec^{2}{x}+\frac{\tan{x}}{x}$$

Here is where you went wrong.
$$\int\frac{\tan{x}}{x}dx=\log{|x|}\tan{x}-\int\log{|x|}\sec^{2}{x}dx$$

 

[DeeAytch]

How about using the Fourier series for tan?

What would this look like?

 

[Goa'uld]

What would this look like?

Here is a derivation someone has already done. It is a .pdf.

http://web.mit.edu/jorloff/www/18.03-esg/notes/fourier-tan.pdf

It converges poorly, however.