Calculate the indefinite integral $\int \frac{x}{\sin x}\,{\rm d}x$

Question

I have to calculate this integral $\int \frac{x}{\sin x}\,{\rm d}x$. Is there any way to evaluate this?

Thanks.

Answer 1

This function's primitive is not expressable in terms of elementary functions. Basically, you're integrating $\frac{1}{\frac{\sin(x)}{x}}$

See here for a detailed explanation

Answer 2

This integral doesn't have a closed-form solution.

See here Wolfram Alpha

Answer 3

The integrand does not admit a closed form antiderivative. See Liouville's theorem and the Risch algorithm for more information. However, its definite counterpart evaluated over $\bigg(0,~\dfrac\pi2\bigg)$ yields twice the value of Catalan's constant as a result.

Answer 4

The integral is not elementary but can be expressed in terms of dilogarithm function as

$$\boxed{\int \frac{x}{\sin x} d x =x \ln \left(\frac{1-e^{i x}}{1+e^{i x}}\right) +i\left[\operatorname{Li_2}\left(-e^{i x}\right)-\operatorname{Li_2}\left(e^{i x}\right)\right]+C \;} $$

By complex number, we transform the integral into $$ \int \frac{x}{\sin x} d x=2 i \int \frac{x}{e^{i x}-e^{-i x}} d x= 2 i \int \frac{x e^{i x}}{e^{2 i x}-1} d x $$ Using integration by parts, we have $$ \int \frac{x}{\sin x} d x =\int x \, d \left(\ln \left(\frac{1-e^{i x}}{1+e^{i x}}\right)\right) =x \ln \left(\frac{1-e^{i x}}{1+e^{i x}}\right)-\int \ln \left(\frac{1-e^{i x}}{1+e^{i x}}\right)d x $$ For the last integral, $$ \int \ln \left(1-e^{i x}\right) d x \stackrel{y=e^{ix}}{=} \int \frac{\ln (1-y)}{-i y} d y= i \operatorname{Li_2}\left(y\right) =i \operatorname{Li_2}\left(e^{i x}\right)+C_1 $$ Similarly, $$ \int \ln \left(1+e^{i x}\right) d x \stackrel{z=-e^{ix}}{=} \int \frac{\ln (1-z)}{i z} d z =-i \operatorname{Li_2}\left(z\right) =-i \operatorname{Li_2}\left(-e^{i x}\right)+C_2 $$ Now we can conclude that $$\int \frac{x}{\sin x} d x =x \ln \left(\frac{1-e^{i x}}{1+e^{ix}}\right) +i\left[\operatorname{Li_2}\left(-e^{i x}\right)-\operatorname{Li_2}\left(e^{i x}\right)\right]+C $$