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

- anonymous

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

- chestercat

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- anonymous

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

- Astrophysics

Try \[x = \tan \theta \]

- anonymous

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

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## More answers

- xapproachesinfinity

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

- xapproachesinfinity

unless i made ARITHMETIC mistake should be good value

- xapproachesinfinity

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

- xapproachesinfinity

lost complex to me to do that integral

- xapproachesinfinity

looks*

- anonymous

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

- xapproachesinfinity

then i must have made arithmetic mistake somewhere

- anonymous

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

- anonymous

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

- xapproachesinfinity

oh right wolfram gives that solution too

- xapproachesinfinity

let me see my math

- anonymous

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

- xapproachesinfinity

oh i see where i made a mistake

- xapproachesinfinity

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

- anonymous

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:
\[ \int_0^{\frac{\pi}{4}} (\ln( \sqrt{2}( \cos \theta \cos \frac{\pi}{4}- \sin \theta \sin \frac{\pi}{4}))- \ln( \cos \theta)) d \theta \\ = \int_0^{\frac{\pi}{4}} ( \frac{1}{2} \ln(2)+ \ln( \cos \theta -\frac{\pi}{4})- \ln( \cos \theta)) d \theta \]

- anonymous

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

- xapproachesinfinity

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

- anonymous

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:
\[ \int_0^{\frac{\pi}{4}} \frac{1}{2} \ln(2) d \theta=\frac{\pi \ln2}{8} \]

- xapproachesinfinity

oh very nice

- xapproachesinfinity

neat work!

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

What was the flaw

- xapproachesinfinity

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

- anonymous

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

- anonymous

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

- xapproachesinfinity

ah yes! some good tricks!

- anonymous

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

- anonymous

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

- xapproachesinfinity

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

- anonymous

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

- Astrophysics

Hey thanks for the tag! Nice question for sure!

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