Menu Close

1-4-ln-x-dx-




Question Number 181840 by srikanth2684 last updated on 01/Dec/22
∫_(−1) ^4 ln x dx
$$\underset{−\mathrm{1}} {\overset{\mathrm{4}} {\int}}{ln}\:{x}\:{dx} \\ $$
Commented by mr W last updated on 01/Dec/22
question is wrong.   for ln (x) to be defined, x>0 !
$${question}\:{is}\:{wrong}.\: \\ $$$${for}\:\mathrm{ln}\:\left({x}\right)\:{to}\:{be}\:{defined},\:{x}>\mathrm{0}\:! \\ $$
Commented by CElcedricjunior last updated on 01/Dec/22
∫_(−1) ^4 lnxdx=[xln∣x∣−x]_(−1) ^4 =+∞
$$\int_{−\mathrm{1}} ^{\mathrm{4}} \boldsymbol{{lnxdx}}=\left[\boldsymbol{{xln}}\mid\boldsymbol{{x}}\mid−{x}\right]_{−\mathrm{1}} ^{\mathrm{4}} =+\infty \\ $$
Commented by Frix last updated on 01/Dec/22
+∞ is wrong!  ∫(dx/x)=ln ∣x∣ +C  In this case the absolute value makes sense  because obviously the area between the  graph of y=(1/x) and the x−axis is the same  in both intervals [a, b] and [−b, −a] ⇒  ∫_(−r) ^(+r) (dx/x)=0∀r∈R despite of the singularity.  ∫ln x dx=−x(1−ln x)+C  In this case y=ln x is not defined for x≤0  and x, y∈R  If we let x, y∈C: x<0∧r=∣x∣ ⇔ x=re^(iπ)  and  ln x =ln r +πi ⇒  ∫_(−1) ^4 ln x dx=−5+8ln 2 +πi  [real (ln x) =ln ∣x∣; imag (ln x) =((π(1−sign x))/2)]
$$+\infty\:\mathrm{is}\:\mathrm{wrong}! \\ $$$$\int\frac{{dx}}{{x}}=\mathrm{ln}\:\mid{x}\mid\:+{C} \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{case}\:\mathrm{the}\:\mathrm{absolute}\:\mathrm{value}\:\mathrm{makes}\:\mathrm{sense} \\ $$$$\mathrm{because}\:\mathrm{obviously}\:\mathrm{the}\:\mathrm{area}\:\mathrm{between}\:\mathrm{the} \\ $$$$\mathrm{graph}\:\mathrm{of}\:{y}=\frac{\mathrm{1}}{{x}}\:\mathrm{and}\:\mathrm{the}\:{x}−\mathrm{axis}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same} \\ $$$$\mathrm{in}\:\mathrm{both}\:\mathrm{intervals}\:\left[{a},\:{b}\right]\:\mathrm{and}\:\left[−{b},\:−{a}\right]\:\Rightarrow \\ $$$$\underset{−{r}} {\overset{+{r}} {\int}}\frac{{dx}}{{x}}=\mathrm{0}\forall{r}\in\mathbb{R}\:\mathrm{despite}\:\mathrm{of}\:\mathrm{the}\:\mathrm{singularity}. \\ $$$$\int\mathrm{ln}\:{x}\:{dx}=−{x}\left(\mathrm{1}−\mathrm{ln}\:{x}\right)+{C} \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{case}\:{y}=\mathrm{ln}\:{x}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for}\:{x}\leqslant\mathrm{0} \\ $$$$\mathrm{and}\:{x},\:{y}\in\mathbb{R} \\ $$$$\mathrm{If}\:\mathrm{we}\:\mathrm{let}\:{x},\:{y}\in\mathbb{C}:\:{x}<\mathrm{0}\wedge{r}=\mid{x}\mid\:\Leftrightarrow\:{x}={r}\mathrm{e}^{\mathrm{i}\pi} \:\mathrm{and} \\ $$$$\mathrm{ln}\:{x}\:=\mathrm{ln}\:{r}\:+\pi\mathrm{i}\:\Rightarrow \\ $$$$\underset{−\mathrm{1}} {\overset{\mathrm{4}} {\int}}\mathrm{ln}\:{x}\:{dx}=−\mathrm{5}+\mathrm{8ln}\:\mathrm{2}\:+\pi\mathrm{i} \\ $$$$\left[\mathrm{real}\:\left(\mathrm{ln}\:{x}\right)\:=\mathrm{ln}\:\mid{x}\mid;\:\mathrm{imag}\:\left(\mathrm{ln}\:{x}\right)\:=\frac{\pi\left(\mathrm{1}−\mathrm{sign}\:{x}\right)}{\mathrm{2}}\right] \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *