2
$\begingroup$

The integral :-

$$\int x^m \ln(a+x) \,dx.$$

(Also what is $m$ is not an integer, just an arbitrary real number?) I have found the integral in the book gradshteyn and ryzhik of which this is a special case. I tried integral by parts for $a=0$ it follows trivially but for the other case please help to find the integral?

$\endgroup$
  • 2
    $\begingroup$ Perhaps you could do $u=a+x$ and then expand $x^m=(u-a)^m$ by the binomial theorem. $\endgroup$ – J_P Jun 21 at 18:28
  • 1
    $\begingroup$ Oh yes thank you very much $\endgroup$ – Bijayan Ray Jun 21 at 18:30
3
$\begingroup$

Use integration by parts and note that $$\frac{x^{m+1}}{a+x}=\sum_{k=0}^m(-a)^kx^{m-k}+\frac{(-a)^{m+1}}{a+x}$$

$\endgroup$
1
$\begingroup$

we have: $$I=\int x^m\ln(a+x)dx$$ now with $u=\ln(a+x)$ we get: $$I=\int(e^u-a)^me^uu\,du$$ now using integration by parts: $$I=\frac{u(e^u-a)^{m+1}}{m+1}-\int\frac{(e^u-a)^{m+1}}{m+1}du$$ now try using binomial expansion. One way of doing it would be by writing: $$(e^u-a)^{m+1}=e^{(m+1)u}(1-ae^{-u})^{m+1}$$

$\endgroup$
1
$\begingroup$

With one step of by-parts integration (on $x^m$), you get rid of the logarithm and reduce to an incomplete Beta integral. https://en.wikipedia.org/wiki/Beta_function (check the fifth property and the incomplete function).

This indirectly proves that for general $m$ there is no closed-form expression.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.