The result should be π/4, you dropped a factor of ½ in the IBP section. Also, how about just performing a magical transformation straight from x to -x, and seeing where that takes you? You'll be pleasantly surprised.
@@GreenMeansGOF After you simplify your transformed integral, you'll be left with log(1+e^-x), which can be re-written as log((1+e^x)/(e^x)). Just apply properties of logarithms, and you'll notice that you're left with your original integral and some other basic Calc 2 integral.
Your solution is more elegant than mine. I IBP like you did but then I brute forced the resultant integral by splitting the integral at 0 and using geometric series for 1/(1+exp(-x)). Had to derive some other general integrals so it took me almost 2 whiteboards to get to the answer XD.
I was an integration by parts hater, until I met this integral... bro this was so awesome! although I dought that I would have been able to guess a factorization of e^x/2... thanks for teaching me this trick!
I think a way to conceptually understand this is by a "limiting integral". We have I = S_-oo^oo x*ln(1+e^x)/(1+x^2)^2 dx Now what is interesting about this integral is that if you take the limit x --> +oo, we can define this to be, since e^x >> 1, IL ~= S_-oo^oo x*ln(e^x)/(1+x^2)^2 dx = S_-oo^oo x*x/(1+x^2)^2 dx = S_-oo^oo [x/(1+x^2)^2]*x dx which we then integrate by parts Differentiate x Integrate [x/(1+x^2)^2]dx --> -1/2(1+x^2)^2 and in the first limit we get IL ~= 0 - (-1/2)*S_-oo^oo dx/(1+x^2)^2 = 1/2 * (pi/2) = pi/4 which is exactly our result. So using the integral version of the squeeze theorem, if IL = lim(I as x-> oo) = pi/4 then so does I at the bounds x --> +-oo since the integral is symmetric.
Integration by parts + odd even decomposition of function odd even decomposition of function Function can be decomposed into f(x) = (f(x)-f(-x))/2 + (f(x)+f(-x))/2 and (f(x)-f(-x))/2 is always odd and (f(x)+f(-x))/2 is always even Finally we get integral 1/2Int(1/(1+x^2),x=0..infinity)
