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I am supposed to solve this integral but i have no idea how:

$$\int_0^1 \frac{\log (x \sqrt{x})}{\sqrt{x}} \,dx$$

Since one limit is $0$ it will be divided by zero.

Can someone please explain this to me (I really want to understand) and guide me through step by step. Thanks! :)

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  • $\begingroup$ So, what integral are we talking about? $\endgroup$
    – Hrodelbert
    Commented Mar 11, 2015 at 9:55
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    $\begingroup$ Do you know the definition of an improper integral (as a limit of proper integrals)? $\endgroup$ Commented Mar 11, 2015 at 9:56
  • $\begingroup$ i think i posted a picture, can't you see it? @Hrodelbert No i don't think so :/ Travis $\endgroup$
    – Louise
    Commented Mar 11, 2015 at 10:02
  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$
    – AlexR
    Commented Mar 11, 2015 at 10:14

3 Answers 3

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Use the transformation $x=z^2$, to get $$I=6\int_{0}^1 \ln z dz $$ and then use integration by parts to obtain $I=-6$.

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  • $\begingroup$ but then i get 6[sqrt(p)*ln(sqrt(p))-sqrt(p)] and that means that i get sqrt(0)*ln(0)-0, am i allowed to do that since ln(0) is undefined? $\endgroup$
    – Louise
    Commented Mar 11, 2015 at 10:54
  • $\begingroup$ Note, $\lim_{x\to 0}x\ln x=0$ $\endgroup$ Commented Mar 11, 2015 at 14:09
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Hint. You know that a primitive of $\log u$ is $u\log u -u$, then using the change of variable $\displaystyle u=\sqrt{x}$, $\displaystyle du=\frac 1{2\sqrt{x}}dx$, you obtain $$ \int_0^1 \frac{\log (x \sqrt{x})}{\sqrt{x}} \,dx= 2 \int_0^1 \log (u^3) \,du=6\times[u\log u -u]_0^1=-6. $$

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  • $\begingroup$ But can i insert 0 in this? ln(0) is undefined $\endgroup$
    – Louise
    Commented Mar 11, 2015 at 11:17
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There are two things I would do first looking at this integral.

The first is resolve the $x\sqrt{x}$ into $x^\frac{3}{2}$, then use the log laws to write $\frac{3}{2}\log(x)$.

This gives you a much nicer log expression, which then makes integration by parts a much easier process.

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