To determine whether the improper integral converges or diverges, we can analyze the behavior of the integrand as x approaches the limits of integration (0 and ∞). If the integral converges, we can evaluate it using numerical integration methods.
We need to evaluate the improper integral
∫0∞e4x+e−4x1dx.
To determine whether this integral converges or diverges, we will analyze the behavior of the integrand as x approaches 0 and ∞.
As x→0, we can approximate the integrand:
e4x≈1ande−4x≈1.
Thus,
e4x+e−4x≈2,
leading to
e4x+e−4x1≈21.
The integral near 0 behaves like
∫0ϵ21dx=2ϵ,
which converges as ϵ→0.
As x→∞, we have
e4x→∞ande−4x→0.
Thus,
e4x+e−4x≈e4x,
leading to
e4x+e−4x1≈e4x1.
The integral behaves like
∫M∞e4x1dx,
which converges since
∫e4x1dx=−4e4x1.
Since the integral converges, we can evaluate it:
∫0∞e4x+e−4x1dx.
Using the substitution u=e4x, we have du=4e4xdx or dx=4udu. The limits change from x=0 to x=∞ which corresponds to u=1 to u=∞:
∫1∞u+u11⋅4udu=41∫1∞u2+11du.
The integral
∫u2+11du=tan−1(u),
evaluated from 1 to ∞ gives:
[tan−1(u)]1∞=2π−4π=4π.
Thus,
∫0∞e4x+e−4x1dx=41⋅4π=16π.
The integral converges and its value is
16π.