What Steps Are Needed to Solve This Integral Problem?

[EngWiPy]

Hi,

I am trying to solve this integral:

$$\int_0^{\infty}\frac{\ln(1+\alpha\,x)}{(1+x)^2}\,dx$$

Using integration by parts this can be written as:

$$-\frac{1}{1+x}\ln(1+\alpha\,x)\Big|_0^{\infty}\Big. + \alpha\int_0^{\infty}\frac{1}{(1+x)(1+\alpha\,x)}\,dx$$

The first term evaluates to zero. The second term can be evaluated using partial fractions as:

$$\int_0^{\infty}\frac{A}{1+x}\,dx+\int_0^{\infty}\frac{\alpha B}{1+\alpha x}\,dx$$

But these integrals aren't finite, are they? A paper I was reading evaluted the original integral to

$$\frac{\alpha}{\alpha-1}\ln(\alpha)$$

and I wonder how did the authors reached this result? Did I do something wrong?

Thanks in advance

 

[kuruman]

Don't use partial fractions. You need to find the right substitution. Hint: Integral (14) here, http://integral-table.com/downloads/single-page-integral-table.pdf

 

[EngWiPy]

Thanks. I searched for a similar formula in the Table of Integrals, Series, and Products, but didn't find one! Applying the formula implies that

$$\lim_{x\to\infty}\,\,\ln\left(\frac{1+x}{\alpha^{-1}+x}\right) = 0$$

Why? How to mathematically show this is the case? Do we need to use the Taylor series expansion?

 

[kuruman]

Why? How to mathematically show this is the case? Do we need to use the Taylor series expansion?

When $x >>(1, ~\alpha^{-1})$, $\frac{1+x}{~\alpha^{-1}+x}\approx\frac{x}{x}=1$. What is $\ln(1)$?

 

[EngWiPy]

When $x >>(1, ~\alpha^{-1})$, $\frac{1+x}{~\alpha^{-1}+x}\approx\frac{x}{x}=1$. What is $\ln(1)$?

Right. $\ln(1) = 0$. Approximations make things easy. Thanks a lot