I've been trying to resolve a calculus question and seem to be having troubles understanding exactly how to approach it. Some hints are supplied, but they don't exactly seem to help. Thanks to anyone who lends a hand!
Without evaluating the integral, show that the $\int_{0}^{\infty} \frac{ln(x)}{(x^2+1)}dx = 0$. (HINTS: let $x = \frac{1}{u}$ in order to re-express the integral in terms of u. Compare the two integrals, and recall the basic fact that the only number equal to its negative is zero.)
If $x = \frac 1 u$, what is $dx = \frac{dx}{du} du$? Substitute this expression in. How do the limits of integration change?
What is $\ln(x)$ in terms of $\ln(u)$?
What happens to the denominator if instead of $x^2 + 1$ we write $(\frac 1 u)^2 + 1$, and then simplify?
In more detail, the hint says to do the following. If $x = \frac 1 u$, then $\frac{dx}{du} = -\frac 1{u^2}$, so we can replace $dx$ with $-\frac{1}{u^2}du$ and the limits of integration $x = 0$ and $x = \infty$ with $\lim_{x \to \infty} u = \lim_{x \to \infty} \frac 1 x = 0$ and $\lim_{x \to 0} u = \lim_{x \to 0} \frac 1 x = \infty$.
Write $I$ for your original integral. Then, at this stage, we have
$$I = \int_\infty^0 \frac{\ln x}{x^2+1}\left(-\frac 1{u^2}\right)\, du = \int_0^\infty \frac{\ln x}{u^2(x^2+1)}du.$$
Here we've switched the upper and lower limits of integration, picking up an extra minus sign in the process, which cancels with the minus sign from $-1/u^2$.
Since $x = u^{-1}$, we also know $\ln x = \ln u^{-1} = - \ln u$. Let's substitute that in. We get
$$I = \int_0^\infty \frac{-\ln u}{u^2(x^2+1)}dx = -\int_0^\infty \frac{\ln u}{u^2(x^2+1)}du.$$
The denominator is $x^2 + 1 = (\frac 1 u)^2 + 1 = \frac{1+u^2}{u^2}$. Let's substitute that in:
$$ I = -\int_0^\infty \frac{\ln u}{u^2\frac{1+u^2}{u^2}}du = -\int_0^\infty \frac{\ln u}{1+u^2}du.$$
But this last thing is $-I$, so we have $I = -I$. That means $2I = 0$, so $I = 0$.