- #1
pierce15
- 315
- 2
Check this out:
[tex]\int_{-1}^{0} ln(x) dx[/tex]
[tex]u=ln(x), dv=dx[/tex]
[tex]du=\frac{1}{x},v=x[/tex]
[tex]\int_{-1}^{0} ln(x) dx= x\space ln(x) \|_{-1}^{0} -\int_{-1}^{0} \frac{x}{x} dx[/tex]
[tex]=x\space ln(x) \|_{-1}^{0}-x \|_{-1}^{0}[/tex]
[tex]=\lim_{x\to0} x\space ln(x)-(-ln(-1))-(0-(-1))[/tex]
[tex]\lim_{x\to0} x\space ln(x)=lim_{x\to0} \frac{ln(x)}{1/x}[/tex]
[tex]=\lim_{x\to0} \frac{1/x}{-1/x^2}[/tex]
[tex]=\lim_{x\to0} -x=0[/tex]
[tex]\int_{-1}^{0} ln(x)dx=ln(-1)-1[/tex]
[tex]e^{i\pi}=-1\to ln(-1)=i\pi[/tex]
[tex]\int_{-1}^{0}ln(x)\space dx=i\pi-1[/tex]
Is this legitimate?
P.S. Why don't my limits look right?
[tex]\int_{-1}^{0} ln(x) dx[/tex]
[tex]u=ln(x), dv=dx[/tex]
[tex]du=\frac{1}{x},v=x[/tex]
[tex]\int_{-1}^{0} ln(x) dx= x\space ln(x) \|_{-1}^{0} -\int_{-1}^{0} \frac{x}{x} dx[/tex]
[tex]=x\space ln(x) \|_{-1}^{0}-x \|_{-1}^{0}[/tex]
[tex]=\lim_{x\to0} x\space ln(x)-(-ln(-1))-(0-(-1))[/tex]
[tex]\lim_{x\to0} x\space ln(x)=lim_{x\to0} \frac{ln(x)}{1/x}[/tex]
[tex]=\lim_{x\to0} \frac{1/x}{-1/x^2}[/tex]
[tex]=\lim_{x\to0} -x=0[/tex]
[tex]\int_{-1}^{0} ln(x)dx=ln(-1)-1[/tex]
[tex]e^{i\pi}=-1\to ln(-1)=i\pi[/tex]
[tex]\int_{-1}^{0}ln(x)\space dx=i\pi-1[/tex]
Is this legitimate?
P.S. Why don't my limits look right?
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