The odd part of this peculiar integral $$f(x)=\int_{0}^x\frac{\cos(z)}{e^{\frac{1}{z}}+1}dz$$ seems to be equal to half the sine function, that is $$f(x)-f(-x)=\sin(x)$$ I have only observed this graphically through Desmos, but could anyone explain why this is true?
0
$\begingroup$
$\endgroup$
5
$\begingroup$
$\endgroup$
$$ f(-x) = \int_0^{-x} \dfrac{\cos(z)\; dz}{\exp(1/z)+1} = -\int_0^x \dfrac{\cos(z)\; dz}{\exp(-1/z)+1} $$ Now note that $$ \dfrac{1}{\exp(1/z)+1} + \frac{1}{\exp(-1/z)+1} = \frac{1}{\exp(1/z)+1} + \frac{\exp(1/z)}{1+\exp(1/z)} = 1 $$
answered Dec 7 '18 at 20:05
3
$\begingroup$
$\endgroup$
$$(f(x)-f(-x))'=\frac{\cos x}{e^{1/x}+1}+\frac{\cos x}{e^{-1/x}+1}=\cos x.$$
