Solveeit Logo
Login

Question

Question: If $I_n = \int_0^\infty e^{-x} (sinx)^n dx (n>1)$, then the value of $\frac{101 I_{10}}{I_8}$ is equ...

If In=∫0∞e−x(sinx)ndx(n>1)I_n = \int_0^\infty e^{-x} (sinx)^n dx (n>1), then the value of 101I10I8\frac{101 I_{10}}{I_8} is equal to ______.

Answer

90

Explanation

Solution

Let In=∫0∞e−x(sin⁡x)ndxI_n = \int_0^\infty e^{-x} (\sin x)^n dx for n>1n>1. We want to find the value of 101I10I8\frac{101 I_{10}}{I_8}.

We use integration by parts to find a reduction formula for InI_n. Let u=(sin⁡x)nu = (\sin x)^n and dv=e−xdxdv = e^{-x} dx. Then du=n(sin⁡x)n−1cos⁡xdxdu = n(\sin x)^{n-1} \cos x dx and v=−e−xv = -e^{-x}. In=[−e−x(sin⁡x)n]0∞−∫0∞(−e−x)n(sin⁡x)n−1cos⁡xdxI_n = \left[ -e^{-x} (\sin x)^n \right]_0^\infty - \int_0^\infty (-e^{-x}) n(\sin x)^{n-1} \cos x dx. The boundary term is lim⁡x→∞(−e−x(sin⁡x)n)−(−e−0(sin⁡0)n)\lim_{x \to \infty} (-e^{-x} (\sin x)^n) - (-e^{-0} (\sin 0)^n). Since ∣sin⁡x∣≤1|\sin x| \le 1, ∣e−x(sin⁡x)n∣≤e−x|e^{-x} (\sin x)^n| \le e^{-x}, and lim⁡x→∞e−x=0\lim_{x \to \infty} e^{-x} = 0, the term at infinity is 0 by the Squeeze Theorem. At x=0x=0, (sin⁡0)n=0(\sin 0)^n = 0 for n>1n>1. So the boundary term is 0−0=00 - 0 = 0. In=n∫0∞e−x(sin⁡x)n−1cos⁡xdxI_n = n \int_0^\infty e^{-x} (\sin x)^{n-1} \cos x dx.

Let's apply integration by parts again to the new integral. Let Jn=∫0∞e−x(sin⁡x)ncos⁡xdxJ_n = \int_0^\infty e^{-x} (\sin x)^n \cos x dx. So In=nJn−1I_n = n J_{n-1}. Consider Jn−1=∫0∞e−x(sin⁡x)n−1cos⁡xdxJ_{n-1} = \int_0^\infty e^{-x} (\sin x)^{n-1} \cos x dx. Let u=(sin⁡x)n−1cos⁡xu = (\sin x)^{n-1} \cos x and dv=e−xdxdv = e^{-x} dx. Then du=[(n−1)(sin⁡x)n−2cos⁡2x−(sin⁡x)n−1sin⁡x]dx=[(n−1)(sin⁡x)n−2(1−sin⁡2x)−(sin⁡x)n]dx=[(n−1)(sin⁡x)n−2−(n−1)(sin⁡x)n−(sin⁡x)n]dx=[(n−1)(sin⁡x)n−2−n(sin⁡x)n]dxdu = [(n-1)(\sin x)^{n-2} \cos^2 x - (\sin x)^{n-1} \sin x] dx = [(n-1)(\sin x)^{n-2} (1-\sin^2 x) - (\sin x)^n] dx = [(n-1)(\sin x)^{n-2} - (n-1)(\sin x)^n - (\sin x)^n] dx = [(n-1)(\sin x)^{n-2} - n(\sin x)^n] dx. And v=−e−xv = -e^{-x}. Jn−1=[−e−x(sin⁡x)n−1cos⁡x]0∞−∫0∞(−e−x)[(n−1)(sin⁡x)n−2−n(sin⁡x)n]dxJ_{n-1} = \left[ -e^{-x} (\sin x)^{n-1} \cos x \right]_0^\infty - \int_0^\infty (-e^{-x}) [(n-1)(\sin x)^{n-2} - n(\sin x)^n] dx. The boundary term is lim⁡x→∞(−e−x(sin⁡x)n−1cos⁡x)−(−e−0(sin⁡0)n−1cos⁡0)\lim_{x \to \infty} (-e^{-x} (\sin x)^{n-1} \cos x) - (-e^{-0} (\sin 0)^{n-1} \cos 0). Similar to the previous boundary term, the limit at infinity is 0. At x=0x=0, (sin⁡0)n−1=0(\sin 0)^{n-1} = 0 for n−1>0n-1 > 0 (i.e., n>1n>1), so the term at x=0x=0 is 0. The boundary term is 0. Jn−1=∫0∞e−x[(n−1)(sin⁡x)n−2−n(sin⁡x)n]dxJ_{n-1} = \int_0^\infty e^{-x} [(n-1)(\sin x)^{n-2} - n(\sin x)^n] dx. Jn−1=(n−1)∫0∞e−x(sin⁡x)n−2dx−n∫0∞e−x(sin⁡x)ndxJ_{n-1} = (n-1) \int_0^\infty e^{-x} (\sin x)^{n-2} dx - n \int_0^\infty e^{-x} (\sin x)^n dx. Jn−1=(n−1)In−2−nInJ_{n-1} = (n-1) I_{n-2} - n I_n.

Substitute Jn−1=InnJ_{n-1} = \frac{I_n}{n}: Inn=(n−1)In−2−nIn\frac{I_n}{n} = (n-1) I_{n-2} - n I_n. Multiply by nn: In=n(n−1)In−2−n2InI_n = n(n-1) I_{n-2} - n^2 I_n. In+n2In=n(n−1)In−2I_n + n^2 I_n = n(n-1) I_{n-2}. (1+n2)In=n(n−1)In−2(1 + n^2) I_n = n(n-1) I_{n-2}. In=n(n−1)n2+1In−2I_n = \frac{n(n-1)}{n^2+1} I_{n-2}. This is the reduction formula for InI_n.

We need to find the value of 101I10I8\frac{101 I_{10}}{I_8}. Using the reduction formula with n=10n=10: I10=10(10−1)102+1I10−2=10⋅9100+1I8=90101I8I_{10} = \frac{10(10-1)}{10^2+1} I_{10-2} = \frac{10 \cdot 9}{100+1} I_8 = \frac{90}{101} I_8.

Now, we can find the ratio I10I8\frac{I_{10}}{I_8}: I10I8=90101\frac{I_{10}}{I_8} = \frac{90}{101}.

Finally, we calculate the required value: 101I10I8=101×(I10I8)=101×(90101)=90\frac{101 I_{10}}{I_8} = 101 \times \left( \frac{I_{10}}{I_8} \right) = 101 \times \left( \frac{90}{101} \right) = 90.

The final answer is 90.