Can differentiation under the integral sign simplify complicated integrals?

[seonguvai]

I happened to notice that if you apply differentiation under the integral sign, then [FONT=MathJax_Size2]∫[FONT=MathJax_Main]∞[FONT=MathJax_Main]0[FONT=MathJax_Math]e[FONT=MathJax_Main]−[FONT=MathJax_Math]a[FONT=MathJax_Math]x[FONT=MathJax_Main]−[FONT=MathJax_Math]e[FONT=MathJax_Main]−[FONT=MathJax_Math]b[FONT=MathJax_Math]x[FONT=MathJax_Math]x[FONT=MathJax_Math]d[FONT=MathJax_Math]x[FONT=MathJax_Main]=[FONT=MathJax_Main]ln[FONT=MathJax_Math]b[FONT=MathJax_Main]−[FONT=MathJax_Main]ln[FONT=MathJax_Math]a[FONT=MathJax_Main]=[FONT=MathJax_Size2]∫[FONT=MathJax_Math]b[FONT=MathJax_Math]a[FONT=MathJax_Main]1[FONT=MathJax_Math]x[FONT=MathJax_Math]d[FONT=MathJax_Math]x
Is this correct? If so is it possible to write other complicated improper integrals as simply as this?

 

[alyafey22]

I think what you wrote is

$$\int^\infty_0 \frac{e^{-ax}-e^{-bx}}{x}\,dx = \ln(b)-\ln(a)$$

differentiation under the integral sign can be used to solve many complicated integrals.

Take a look at my http://mathhelpboards.com/calculus-10/advanced-integration-techniques-3233.html.