To solve the limit of the given integral as m→∞, we first need to analyze the integrand. The integral is:
lim∫−∞∞1+x2+x4+⋯+x2mdx.
Step 1: Simplifying the Integrand
The integrand can be expressed as:
1+x2+x4+⋯+x2m1.
For large values of x, the term x2m will dominate the sum 1+x2+x4+⋯+x2m. Therefore, for large x, the integrand can be approximated by:
x2m1.
For small values of x, specifically ∣x∣<1, the sum 1+x2+x4+⋯+x2m approaches a finite value as m→∞. We need to consider these two cases separately.
Step 2: Asymptotic Behavior Analysis
Case 1: ∣x∣>1
For ∣x∣>1:
1+x2+x4+⋯+x2m≈x2m,
so the integrand approximates to:
x2m1.
As m→∞, x2m1 tends to 0 for any fixed x with ∣x∣>1.
Case 2: ∣x∣<1
For ∣x∣<1:
1+x2+x4+⋯+x2m≈1−x21−x2m+2.
As m→∞, x2m+2→0, so:
1+x2+x4+⋯+x2m≈1−x21.
Thus, the integrand approximates to:
11−x2.
Step 3: Integral Evaluation
As m→∞, for ∣x∣<1, the integrand behaves as 1−x2. However, for ∣x∣>1, the integrand approaches 0. Thus, the main contribution to the integral comes from the region where ∣x∣<1.
We integrate 1−x2 over the interval −1<x<1:
∫−∞∞1+x2+x4+⋯+x2mdx≈∫−11(1−x2)dx.
Step 4: Computing the Integral
Now, we calculate:
∫−11(1−x2)dx.
This can be split into two simpler integrals:
∫−111dx−∫−11x2dx.
Evaluating each part:
∫−111dx=x∣∣−11=1−(−1)=2, ∫−11x2dx=3x3∣∣−11=313−3(−1)3=31−(−31)=32. Combining these results:
∫−11(1−x2)dx=2−32=36−32=34.
Conclusion
The value of the limit of the integral as m→∞ is:
34.
