Question

evaluate the integral

Answer 1

I actually have no idea, lol. by parts will not work I don't think

Answer 2

Are you sure that the lower limit is zero. ln(0) is -Infinity

Answer 3

Unless this is a Reimann Sum?

Answer 4

there must be problem with limit, ithink

Answer 5

This integral is actually equal to \[- \frac \pi 4
\]

Answer 6

this is also what wolframalpha is giving How to find it out?

Answer 7

type in the function.

Answer 8

Is this a problem in Calculus?

Answer 9

can you give me a hint. I am not able to do it by parts. I

Answer 10

im thinking of substitution \(x = \tan u\)

Answer 11

that gives you a nice bounds and you allows you to use properties of definite integrals

Answer 12

or is this question from complex analysis ?

Answer 13

\[\int_0^{\infty}\left[\frac{1}{(x^2+1)^2}\cdot \ln(x)\right]dx\]Is that how you thought of making \(x=\tan u\)?

Answer 14

I tried a couple things and this is what works! \[\Large I(\alpha) = \int\limits_0^\infty \frac{\ln(\alpha x)}{(x^2+1)^2}dx\] You need to use a parameter, then take the derivative with respect to it to get: \[\Large I'(\alpha) = \int\limits_0^\infty \frac{1}{\alpha} \frac{1}{(x^2+1)^2}dx\] This integral (which is hard but not impossible anymore becomes:

Answer 15

\[\Large I'(\alpha) = \frac{1}{2\alpha} \left( \frac{x}{x^2+1} + \tan^{-1}x\right) |_0^\infty\] Evaluate on the bounds and you get \[\Large I'(\alpha) = \frac{1}{\alpha} \frac{\pi}{4}\] Now let's integrate with respect to alpha: \[\Large I(\alpha) = \ln(\alpha) \frac{\pi}{4} + C\]

Answer 16

To solve for C we set them equal to each other \[\Large I(\alpha) = \ln( \alpha) \frac{\pi}{4} + C = \int\limits_0^\infty \frac{\ln(\alpha x)}{(x^2+1)^2}dx\] And actually I have to go unfortunately but we just need to set alpha so we can solve for the constant which will =0 and then we set alpha to 1 to recover our original integral I believe...

Answer 17

I actually didn't finish solving it yet but I just assumed it was right when I saw the pi/4 but I might have messed up somewhere. Good luck haha.

Answer 18

Ok I'm back let me finish!

Answer 19

Here's where we left off: \[\Large I(\alpha) = \ln( \alpha) \frac{\pi}{4} + C = \int\limits_0^\infty \frac{\ln(\alpha x)}{(x^2+1)^2}dx\] Let's let alpha =e and alpha =1/e to get two equations: \[\Large \frac{\pi}{4} + C = \int\limits\limits_0^\infty \frac{\ln( x)}{(x^2+1)^2}dx \\ \Large -\frac{\pi}{4} + C = -\int\limits\limits_0^\infty \frac{\ln( x)}{(x^2+1)^2}dx\] Add these two together and we have simply \[\Large C=0\]\

Answer 20

Upon looking at it again, I have decided that my answer is wrong. I don't know how to fix it haha.

Answer 21

this parameter stuff is really a very clever method xD

Answer 22

adding them together gives C = -pi/4

Answer 23

@ganeshie8 are you sure? I think I discovered the mistake in this substitution

Answer 24

See we can rewrite this "substitution" as:\[\large I(\alpha) = \int\limits_0^\infty \frac{\ln(\alpha x)}{(x^2+1)^2}dx = \ln(\alpha) \int\limits_0^\infty \frac{1}{(x^2+1)^2}dx+\int\limits_0^\infty \frac{\ln( x)}{(x^2+1)^2}dx\] So there's not really any dependence of the integral on alpha at all.

Answer 25

Oh im getting C = integral,
why is that "independence" a reason for C getting equal to the integral in the end ?

Answer 26

nvm i was working on working on wrong stuff
so we dont have C yet ?

Answer 27

Here is a solution using earlier trig substitution :
\[\begin{aligned}
&\int\limits_0^{\infty} \frac{\ln x dx}{(x^2+1)^2} &&\stackrel{u=\tan x}{=}
\int\limits_0^{\pi/2}\cos^2 u \ln (\tan u) du = I\\~\\
&&&= -\int\limits_0^{\pi/2}\cos^2(\frac{\pi}{2}- u) \ln (\tan (\frac{\pi}{2}-u)) du \\~\\

&&&= -\int\limits_0^{\pi/2}\sin^2 u \ln (\tan u) du = I\\~\\
\end{aligned}\]
Adding both gives
\[\begin{aligned}

2I&= \int\limits_0^{\pi/2}\cos(2u) \ln (\tan u) du \\~\\
\end{aligned}\]

Evaluating the indefinite integral is trivial by parts
\[\begin{aligned}

2I&= \frac{1}{2}\sin(2u) \ln (\tan u) \Bigg|_0^{\pi/2} - \int \limits_0^{\pi/2} 1 du \\~\\
& = -\frac{\pi}{2}
\end{aligned}\]

Answer 28

Suppose we try letting \(u=\ln x\) so that \(e^u=x\), and \(dx=e^u~du\). Then the integral becomes
\[\int_{-\infty}^\infty \frac{ue^u}{(e^{2u}+1)^2}~du\]
which looks ... easier? It seems doable with IBP at first glance.

Answer 29

IBP yields something like
\[\frac{1}{2}\left[\arctan e^u+\frac{u e^u}{e^{2u}+1}\right]_{-\infty}^\infty-\frac{1}{2}\int_{-\infty}^\infty \left(\arctan e^u+\frac{e^u}{e^{2u}+1}\right)~du\]
which runs into the problem of
\[\int_{-\infty}^\infty \arctan e^u~du\]
not converging.

Answer 30

Using a little residue theory one can show that (for \(a>0\))

\[\int\limits_{0}^{\infty}\frac{\ln(x)}{(x^2+a^2)^2}dx=\frac{\pi}{4a^3}\ln\left(\frac{a}{e}
\right)\]

letting \(a=1\) gives the answer to this particular problem