# $\int \frac{\sin x}{x}dx=\sin x \log x$?

I know that integration of $$\int \frac{\sin x}{x}dx=\text{Si} (x)$$, but when I tried it myself I am getting a different answer. Please check where I am going wrong.

To find: Closed form of $$\int \dfrac{\sin x}{x}dx$$

Integrating by parts:

$$\int \frac{\sin x}{x}dx=\sin x \log x- \int \cos x \log x dx$$

Now finding: $$\int \cos x \log x$$

$$\int \cos x \log x=-\log x \sin x - \int \frac{-\sin x}x dx$$

Putting value of $$\int \cos x \log x$$ back:

$$\int \frac{\sin x}{x}dx = \sin x \log x- \left(-\log x \sin x - \int \frac{-\sin x}x dx\right)$$

$$2\int \frac{\sin x}{x}dx=2\sin x \log x$$

$$\int \frac{\sin x}{x}dx=\sin x \log x$$

In your second integration by parts, you got the signs wrong (and you forgot a $$dx$$). It should be $$\int \cos x \log x dx = \log x \sin x - \int \frac{\sin x}{x} dx$$ If you plug this back into your first formula, you get: $$\int \frac{\sin x}{x} dx = \log x \sin x - \left( \log x \sin x - \int \frac{\sin x}{x} dx \right) = \int \frac{\sin x}{x} dx$$ which doesn't say much. Basically you did the same integration by parts twice, once in one direction then in reverse. You cannot compute anything like that.
Note that $$\int \cos x \log x=\log x \sin x - \int \frac{\sin x}x dx,$$ as integration of $$\cos x$$ is $$\sin x$$. So your answer was wrong.
Try a different method. From Maclurins series we have $$\sin x=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}.$$ Thus $$\int\frac{\sin x}{x}=\sum_{n=0}^{\infty}\int\frac{(-1)^n x^{2n}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)(2n+1)!}+c.$$