# Whats the Integral of $\int_{-1/2}^{1/2} \cos(x)\ln(\frac{1+x}{1-x})\,dx$

I currently don't see how to solve the following integral:

$$\int_{-1/2}^{1/2} \cos(x)\ln\left(\frac{1+x}{1-x}\right) \,dx$$

I tried to solve it with integration by parts and with a Taylor series, but nothing did help me so far.

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## 4 Answers

As pointed out, the integrand is odd. To make this clear, let $x\to -x:$

$$I=\int_{-1/2}^{1/2} \cos x\ln\left(\frac{1+x}{1-x}\right)dx=\int_{-1/2}^{1/2} \cos x\ln\left(\frac{1-x}{1+x}\right)dx$$

Adding both:

$$I=\frac{1}{2}\int_{-1/2}^{1/2} \cos x\ln\left(1\right)dx=0$$

For future reference, whenever you see symmetry in the bounds, check whether the integrand is even/odd before embarking on anything else!

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Thanks for your answer! Unfortunately I can not see how you get from the first to the second line. (How you add the $\ln$'s.) –  leo Jan 23 '13 at 14:50
Factor $\cos x$ and $\ln x+\ln y=\ln xy$ (alternatively, you can revert the fraction in one of them and add a minus sign ($\ln a/b =-\ln b/a$) –  L. F. Jan 23 '13 at 14:52
Ah yes I see! Brilliant! –  leo Jan 23 '13 at 14:57

Hint: The integrand is an odd function.

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You can't solve the indefinite integral using elementary functions. This integral can be solved by parts however if you use $Ci(x)$ - the Cosine Integral. The actual value as @Psx mentioned is of course zero since the integrand is odd.

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Letting $x=-u$ in the second integral you're done $$\int_{-1/2}^{1/2} \cos(x)\ln(1+x) \ dx-\int_{-1/2}^{1/2} \cos(x)\ln(1-x) \ dx$$

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