Evaluating limit by converting to integral

[manankhurma]

Trying to evaluate the a limit, the following integral came up:

\int_0^1 \dfrac{log(1+t)}{t}dt

How can this integral be evaluated using basic techniques?

 

[tiny-tim]

welcome to pf!

hi manankhurma! welcome to pf!

hmm … looks symmetric, doesn't it?

try integrating by parts, then using a substitution, and fiddling around …

what do you get? ​

 

[Nebuchadnezza]

I would use a different method, this is a complicated integral. And it is hard if not impossible to solve it using standard methods. Below is one way to evaluate it

I = \int_{0}^{1} \frac{\log(t+1)}{t} \mathrm{d}t

Since we are integrating over [0,1] a smart idea, is to use the maclaurin expansion of \log(t+1) = t - \frac{1}{2}t^2 + \frac{1}{3}t^3 - \frac{1}{4}t^4 + ... + \frac{1}{n}(-t)^{n}

<< Rest of solution deleted by Moderator >>

 

[DivisionByZro]

Although you did give an answer without the OP showing much effort, I must say that the above solution is very elegant!

 

[berkeman]

I would use a different method, this is a complicated integral. And it is hard if not impossible to solve it using standard methods. Below is one way to evaluate it

I = \int_{0}^{1} \frac{\log(t+1)}{t} \mathrm{d}t

Since we are integrating over [0,1] a smart idea, is to use the maclaurin expansion of \log(t+1) = t - \frac{1}{2}t^2 + \frac{1}{3}t^3 - \frac{1}{4}t^4 + ... + \frac{1}{n}(-t)^{n}

<< Rest of solution deleted by Moderator >>

Even though this thread was originally posted in the general technical math forums, you should assume that the OP is posting something related to schoolwork, especially if they show no effort to solve it on their own.

The thread is moved to the Homeowork Help forums, and I've deleted the end of your post with the worked-out solution. Your initial hint is very good.