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

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?

Homework Helper
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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.

Homework Helper
Then stop with some finite power for an approximation.

Homework Helper
Dearly Missed
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.