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 $$\int\frac{\sin x}{x}=\sum_{n=0}^{\infty}\int\frac{(-1)^n x^{2n}}{(2n+1)!}=\sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)(2n+1)!}+c.$$


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