Solution:
∫y−11dxfrom5 to ∞
can be written as
a→∞lim∫5ay−11dy
y−1=t⟹dy=dty→5;t→4 and y→∞;t→∞∴a→∞lim∫4at1dta→∞lim[[−21+1t−21+1]a−[−21+1t−21+1]4]
a→∞lim[2a−24]a→∞lim[2a−4] As a tends to infinite, entire expression tends to infinite.
Thus the given integral is divergent.
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