# Convergence proof(integral)

#### Also sprach Zarathustra

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
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

$$\displaystyle \chi$$ $$\displaystyle \sigma$$

#### Also sprach Zarathustra

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!