1. ## Integration proof

Which substitution can I use to help me prove that the integral of ln (x) / (1 + x^2) between 0 and 1 is identical to -1 * the same integral between 1 and infinity?

2. Originally Posted by kevinlightman
Which substitution can I use to help me prove that the integral of ln (x) / (1 + x^2) between 0 and 1 is identical to -1 * the same integral between 1 and infinity?
The first thing I notice is that the substitution u= 1/x changes x= 0 to 1 into u= infinity to 1. du= $\displaystyle -\frac{1}{x^2}dx$ so $\displaystyle dx= -x^2 du= -\frac{1}{u^2}du$. The only "tricky" part is recognizing the $\displaystyle x^2+ 1= \frac{1}{u^2}+ 1= \frac{u^2+ 1}{u^2}$ and that is not all that "tricky"!

3. Originally Posted by HallsofIvy
The first thing I notice is that the substitution u= 1/x changes x= 0 to 1 into u= infinity to 1. du= $\displaystyle -\frac{1}{x^2}dx$ so $\displaystyle dx= -x^2 du= -\frac{1}{u^2}du$. The only "tricky" part is recognizing the $\displaystyle x^2+ 1= \frac{1}{u^2}+ 1= \frac{u^2+ 1}{u^2}$ and that is not all that "tricky"!
Great thanks, you're right it does seem a lot less tricky now I can see it working

4. Sorry edited double post