The question is: using the comparison test for improper integrals state if ∫0∞x4e−xdx converges or not.

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The question is: using the comparison test for improper integrals state if  ∫0∞x4e−xdx converges or not.

Neil Dismukes · Accepted Answer

There is some M large enough so that  |x6ex|&amp;lt;1  for x&amp;gt;M, since by LHopitals the limx→0x6ex=0  Then, by monotonicity of integration, we have  ∫M∞e−xx4dx≤∫M∞e−xx−2dx&amp;lt;∞  Of course,  ∫0Me−xx4dx  poses no problem as the integrand is bounded and the interval is finite. Note also you could integrate this by parts without too much trouble and evaluate the integral directly.

Philip Williams · Accepted Answer

∫0∞x4e−xdx=∫0Mx4e−xdx+∫M∞x4e−xdx  ≤∫0Mx4e−xdx+∫M∞x4x−6dx  where e−x&amp;lt;x6 for all x≥M

karton · Accepted Answer

In explicit terms, the convexity inequality ex≥1+x implies that for any ≥0 we have ex/6≥1+x6 and ex≥(1+x6)6. In particular 0≤∫0+∞x4e−xdx≤∫0+∞x4(1+x6)6dx=65∫0+∞z4dz(1+z)6=655 This argument proves n!≤(n+2)n+1n+1. One may prove much better, like ∫0+∞xne−xdx≤(n+1)n+1en by exploiting the log-convexity of the LHS with respect to the n-variable

The question is: using the comparison test for improper integrals

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