$\displaystyle \int^{\infty}_{-\infty}x^{-1/2}e^{x/2}dx $
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Edit: sorry it was wrong
Hello, This is weird because it is not defined for x<0...
Originally Posted by wantanswers $\displaystyle \int^{\infty}_{-\infty}x^{-1/2}e^{x/2}dx $ Well intuitively when say x>1, $\displaystyle e^{x/2}$ grow faster then $\displaystyle \sqrt{x}$ . It follows that $\displaystyle \int_1^\infty \frac{e^{x/2}}{\sqrt{x}}\,dx$ diverges and hence so is the original integral.
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