Convergence proof(integral)

Dec 2009
1,506
434
Russia
I have a problem with proving that the next integral is converges:

\(\displaystyle \int^0_{-1}\frac{ ln(1+t)}{t}dt\)


Thanks!
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
994
near Piacenza (Italy)
You can start from the general result [which of course can be demostrated]...

\(\displaystyle \displaystyle \int_{0}^{1} \tau^{n}\ \ln \tau\ d\tau = -\frac{1}{(n+1)^{2}}\) (1)

... and then performing the substitution \(\displaystyle 1+t=\tau\) You obtain...

\(\displaystyle \displaystyle \int_{-1}^{0} \frac{\ln (1+t)}{t}\ dt = - \int_{0}^{1} \frac{\ln \tau}{1-\tau}\ d\tau = - \sum_{n=0}^{\infty} \int_{0}^{1} \tau^{n}\ \ln \tau\ d\tau = \)

\(\displaystyle \displaystyle = \sum_{n=0}^{\infty} \frac{1}{(n+1)^{2}} = \frac{\pi ^ {2}}{6} \) (2)

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
Dec 2009
1,506
434
Russia
Oh!

Great! (This result I need to prove later actually)

I have tried to do something like this :

\(\displaystyle \int^0_{-1}\frac{ln(1+t))}{t}dt<-2\int^1_0 ln(1+t)dt= -2[(t+1)ln(t+1)-(t+1)]^1_0<\infty\)
Now, by comparison test...

Is that prove the convergence?


Thank you again!