Integrate [itex]\int_0^1\frac{\sin(\pi x)}{1-x}dx[/itex]

[funcalys]

Homework Statement

$\int_0^1\frac{\sin(\pi x)}{1-x}dx$

Homework Equations

$\int \frac{\sin (\pi x)}{1-x}=Si(\pi-\pi x)$

The Attempt at a Solution

I was stuck on the above integral while solving an exercise, I found out earlier on Wolfram that this integral doesn't probably have an elementary antiderivative representation so I cannot just calculate it directly by Newton-Leibniz formula. Integration by parts is not very viable in this case also. Maybe this needs the notion of contour integration to be solved? Any idea guys?

 

[haruspex]

Have you tried expanding the denominator as a power series?

 

[funcalys]

Yeah, looks like it yielded some positive results finally , thanks very much for your help.

 

[funcalys]

Oh, I did expand the denominator as a power series and then integrate term by term by resulting series, but I can't proceed to the next step. The general formula for arbitrary k is too complicated for me.

 

[HallsofIvy]

Then stop with some finite power for an approximation.

 

[Ray Vickson]

Oh, I did expand the denominator as a power series and then integrate term by term by resulting series, but I can't proceed to the next step. The general formula for arbitrary k is too complicated for me.

If you change variables to t = 1-x you will get the value of the integral as $I = \text{Si}(\pi),$ where
$$\text{Si}(x) = \int_0^x \frac{\sin(t)}{t} \, dt.$$ You can very easily expand $\sin(t)$ in a power series and get a very straightforward series for $\text{Si}(x)$ in powers of x. That will give you a simple-enough series for I in powers of $\pi$.

 

[funcalys]

Thank HallsofIvy.

If you change variables to t = 1-x you will get the value of the integral as $I = \text{Si}(\pi),$ where
$$\text{Si}(x) = \int_0^x \frac{\sin(t)}{t} \, dt.$$ You can very easily expand $\sin(t)$ in a power series and get a very straightforward series for $\text{Si}(x)$ in powers of x. That will give you a simple-enough series for I in powers of $\pi$.

This seems like the most plausible way to handle that imo . Thanks.

 

[Ray Vickson]

Thank HallsofIvy.

This seems like the most plausible way to handle that imo . Thanks.

Numerical integration methods such as Simpson's rule should work well also.