# How do you evaluate the integral #int e^x/(1+e^(2x))dx# from #-oo# to #oo#?

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To evaluate the integral ( \int \frac{e^x}{1 + e^{2x}} , dx ) from ( -\infty ) to ( \infty ), we can use the method of residues from complex analysis.

First, consider the function ( f(z) = \frac{e^z}{1 + e^{2z}} ). This function has a simple pole at ( z = \frac{\pi i}{2} ), which corresponds to the real axis when ( x ) is considered.

To find the residue at this pole, we take the limit as ( z ) approaches ( \frac{\pi i}{2} ) of ( (z - \frac{\pi i}{2})f(z) ). After simplification, we find that the residue is ( \frac{1}{2} ).

Since the pole is in the upper half-plane and the contour integral is taken over the entire real line, we use the upper half of a semi-circle contour in the complex plane, with a large radius ( R ). The integral along the semi-circle vanishes as ( R ) approaches infinity.

By the residue theorem, the integral along the real line is equal to ( 2\pi i ) times the residue at ( z = \frac{\pi i}{2} ), which is ( \pi i ).

Thus, ( \int_{-\infty}^{\infty} \frac{e^x}{1 + e^{2x}} , dx = \pi i ).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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