## anonymous one year ago Putnam Calculus Problem: $\int\limits_0^1 \frac{\ln(x+1)dx}{x^2+1}= \ ?$

1. 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

2. Astrophysics

Try $x = \tan \theta$

3. 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

4. xapproachesinfinity

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

5. xapproachesinfinity

unless i made ARITHMETIC mistake should be good value

6. xapproachesinfinity

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

7. xapproachesinfinity

lost complex to me to do that integral

8. xapproachesinfinity

looks*

9. anonymous

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

10. xapproachesinfinity

then i must have made arithmetic mistake somewhere

11. 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

12. anonymous

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

13. xapproachesinfinity

oh right wolfram gives that solution too

14. xapproachesinfinity

let me see my math

15. 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

16. xapproachesinfinity

oh i see where i made a mistake

17. xapproachesinfinity

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

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

19. anonymous

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

20. xapproachesinfinity

21. 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: $\int_0^{\frac{\pi}{4}} \frac{1}{2} \ln(2) d \theta=\frac{\pi \ln2}{8}$

22. xapproachesinfinity

oh very nice

23. xapproachesinfinity

neat work!

24. 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

25. 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

26. anonymous

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

27. anonymous

What was the flaw

28. 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

29. 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

30. 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

31. xapproachesinfinity

ah yes! some good tricks!

32. 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

33. anonymous

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

34. xapproachesinfinity

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

35. 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

36. Astrophysics

Hey thanks for the tag! Nice question for sure!