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Comparison test says that if bigger function is convergent then smaller one must be convergent.But here in this example it doesn't work and I want to know why?

$1/(e^x)$ is bigger or equal to $1/(e^x+1)$ ( between zero and infinite) Improper integral $\int_0^\infty \frac{1}{(e^x)} dx$ is convergent and it is $1$ however, improper integral $\int_0^\infty\frac{ 1}{ (e^x+1)}dx $ is divergent.

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  • $\begingroup$ The second integral is not divergent. It is equal to $ln(2)$. Please check your computations. $\endgroup$
    – maq
    May 26 '17 at 7:57
  • $\begingroup$ thanks but :( i cant understand how , because there is inf-inf . lim(t->inf)[t-ln(1+e^t)+ln(2)] $\endgroup$
    – umut piri
    May 26 '17 at 8:23
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One has $$ 0<\frac1{e^x+1}<\frac1{e^x},\qquad x\ge0, $$ giving $$ 0<\int_0^\infty\frac1{e^x+1}\:dx<\int_0^\infty\frac1{e^x}\:dx=\lim_{M\to\infty}\left[-e^{-x} \right]_0^M=\color{red}{-0}+1<\infty $$ and one may observe that $$ \int_0^\infty\frac1{e^x+1}\:dx=\lim_{M\to\infty}\int_0^M\frac{e^{-x}}{1+e^{-x}}\:dx=\lim_{M\to\infty}\left[-\ln(1+e^{-x}) \right]_0^M=\color{red}{-0}+\ln 2. $$

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