Question

Putnam Calculus Problem: \[\int\limits_0^1 \frac{\ln(x+1)dx}{x^2+1}= \ ?\]

Answer 1

Got another fun one for you fellow math nerds out there.... much more straightforward than the last and this time no silly errors :D

Answer 2

Try \[x = \tan \theta \]

Answer 3

That works @Astrophysics but it takes more than that... I just put this up for a fun challenge to whoever wants to try... if you have a solution by all means post it please :D

Answer 4

IBP works fine i got \(\pi\frac{\ln 2}{2}-\frac{\pi^2}{32}\)

Answer 5

unless i made ARITHMETIC mistake should be good value

Answer 6

if we did x=tanu
we would have to integrate \[\int \ln (\tan u+1)du\]

Answer 7

lost complex to me to do that integral

Answer 8

looks*

Answer 9

Close its \[\pi \frac{ln(2)}{8}\]

Answer 10

then i must have made arithmetic mistake somewhere

Answer 11

Ok give me a sec Ill post my solution but I just put a microwave dinner in and the plastic has melted so now my apartment reeks of plastic hold on

Answer 12

Im interested in your IBP solution though could you post it plz I didnt treat it like that myself

Answer 13

oh right wolfram gives that solution too

Answer 14

let me see my math

Answer 15

Dinner ruined :/ but at least I found another math's enthusiast :D... Ok so I started out with the same substitution yieldling:
\[\int_0^{\frac{\pi}{4}} \ln( \tan \theta +1) d \theta= \int_0^{\frac{\pi}{4}} (\ln( \sin \theta + \cos \theta)- \ln( \cos \theta)) d \theta\]
via factoring out the cos and reduing the log

Answer 16

oh i see where i made a mistake

Answer 17

oh interesting it crossed my mind to proceed that way but felt unsafe

Answer 18

Then, the thing that took me forever to see was this:
\[ \cos \theta + \sin \theta = \sqrt{2}( \cos \theta \cos \frac{\pi}{4}+ \sin \theta \sin \frac{\pi}{4})\] Which yields:
\[ \int_0^{\frac{\pi}{4}} (\ln( \sqrt{2}( \cos \theta \cos \frac{\pi}{4}- \sin \theta \sin \frac{\pi}{4}))- \ln( \cos \theta)) d \theta \\ = \int_0^{\frac{\pi}{4}} ( \frac{1}{2} \ln(2)+ \ln( \cos \theta -\frac{\pi}{4})- \ln( \cos \theta)) d \theta \]

Answer 19

darn that should be:
\[ \int_0^{\frac{\pi}{4}} ( \frac{1}{2} \ln(2)+ \ln( \cos( \theta -\frac{\pi}{4}))- \ln( \cos \theta)) d \theta \]

Answer 20

oh IBP won't work actually had made a severe mistake

Answer 21

Ok so since cosine is an even function and the integration region is [0,pi/4] for the last two terms... those integrals are equal and opposite so they cancel. Thus:
\[ \int_0^{\frac{\pi}{4}} \frac{1}{2} \ln(2) d \theta=\frac{\pi \ln2}{8} \]

Answer 22

oh very nice

Answer 23

neat work!

Answer 24

You can show the cosine cancellation explicitly by simply breaking up the integral and just using a simple u sub to remove the - pi/4 from the argument of cosine

Answer 25

Thank you :D.... I had to sleep on it to get it to work for the life of me I was stuck playing around with various trig identities for a while

Answer 26

Anyways your IBP must not have been that off... you were pretty close except for that pi^2 correction term

Answer 27

What was the flaw

Answer 28

i had int ( arctan/x+1)dx by blind error i made it (arctanx/x^2+1)
but again arctanx/x+1 is not an easy one

Answer 29

ahhh yes I went down that path also... I was getting really annoyed because these are supposed to be solvable without reference to integral tables and arctan(x)/(x+1) is not fun to work with an honestly I had to look up the integral of arctan its been that long

Answer 30

So that's when I just went to bed and I tried again tonight and sure enough product to sum trick to coax out that symmetry

Answer 31

ah yes! some good tricks!

Answer 32

Anyways I got some more hw to do as is... so Im signing off for tonight.. I will probably find and post another as soon as I finish my tests this week cya later :D

Answer 33

@Astrophysics solution is posted if you wanted to have a look but I am closing the question now

Answer 34

yea see ya man! that was some a good problem
i should have given it more thought hhhh
good night

Answer 35

You too... dont sweat it the first one I posted I blew really badly... I solved it in about 5 mins and didnt look back and notice I failed to properly distribute a minus sign (facepalm) no wonder it was so easy

Answer 36

Hey thanks for the tag! Nice question for sure!