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I am new to sine and cosine integrals and I am just doing this out of hobby interest, so please forgive me if this sounds obvious or easy.

I am trying to simplify or solve this expression:

$$\int\frac{\sin(ax)}{x}dx$$

Which looks a lot like the $Si(z)$ integral but with an extra $a$. In my case, $a \neq 1$ so I cannot use the $Si(z)$.

This expression is a part of a large one where I looked at the trig integrals and found the one that matched my needs

$$\int\frac{\cos(ax)}{x^n}dx = -\frac{cos(ax)}{(n - 1)x^{n-1}}-\frac{a}{n-1}\int\frac{\sin(ax)}{x}dx$$

Where in my case $n=2$ and $a$ is a function that depends on $a(x,y)$ and $x$ depends on $x(r, z)$

Where $$a(x,y) = x(y-\frac{1}{C}), C \in \mathbb{R}$$ $$x(r, z) = \sqrt{r^2+z^2}$$

Where $C$ is just a constant.

The question is: how do I express the sine integral in terms of elementary functions like the other parts of the expression? If not, then how do I work with it/use it properly?

J. W. Tanner
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Mathaholic
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    You can change variables, letting $z=ax$, to get this into the more standard form. – lulu Jun 10 '24 at 14:52
  • @lulu Okay, I get this now $$\int\frac{a\sin(z)}{z}dz$$ But I still not sure how to solve it or convert it to elementary functions it to since $a\neq1$. Edit: Do I just use it as the $Si(z)$ function and just take the product with $a$? – Mathaholic Jun 10 '24 at 15:02
  • The Sine Integral can't be expressed in terms of elementary functions, and the value of $a$ doesn't matter much (well, unless it is $0$ of course). – lulu Jun 10 '24 at 15:05
  • @lulu I made an edit to my comment, I said "Do I just use it as the $Si(z)$ function and just take the product with $a$?"

    So I can just evaluate the $Si(z)$ and then multiply by $a$?

     – Mathaholic Jun 10 '24 at 15:07
  • @VincentBatens I think it may answer my question depending on the response from Lulu. Noob here... – Mathaholic Jun 10 '24 at 15:09
  • @Mathaholic the answer to your question is yes, it works like any other variable change – Vincent Batens Jun 10 '24 at 15:13
  • @VincentBatens Thank you, I closed my question. – Mathaholic Jun 10 '24 at 15:14
  • It's like any substitution. But check the details. If $z=ax$ then $dz=a,dx$ so $dx=\frac {dz}a$, making the integral $\int \frac {\sin z}{ax},dz=\int \frac {\sin z}{z}, dz$. Again, we are assuming that $a\neq 0$. – lulu Jun 10 '24 at 15:16

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