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$$I=\int_{0}^{\infty}e^{-nx}\sin^{2k}xdx={(2k)!\over n\prod_{j=1}^{k}(n^2+4j^2)}\tag1$$

Recall

$$\sin^{2k}(x)={1\over 2^{2k}}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\cos[(2k-2j)x]$$

$$I={1\over 2^{2k}}{2k\choose k}\int_{0}^{\infty}e^{-nx}dx+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\int_{0}^{\infty}e^{-nx}\cos[(2k-2j)x]dx\tag2$$

$$I={1\over 2^{2k}n}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\int_{0}^{\infty}e^{-nx}\cos[(2k-2j)x]dx\tag3$$

$$\int_{0}^{\infty}e^{-nx}\cos[(2k-2j)]dx={n\over n^2+(2k-2j)^2}$$

$$I={1\over 2^{2k}n}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\cdot{n\over n^2+(2k-2j)^2}\tag4$$

How can I simplify the LHS sum to the RHS product?

$${1\over 2^{2k}n}{2k\choose k}+{2\over 2^{2k}}\sum_{j=0}^{k-1}(-1)^{k-j}{2k\choose j}\cdot{n\over n^2+(2k-2j)^2}={(2k)!\over n\prod_{j=1}^{k}(n^2+4j^2)}\tag5$$

gymbvghjkgkjkhgfkl
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2 Answers2

4

Olivier Oloa already indicated two approaches in the comments: one by repeated integration by parts (worked out in detail in the answer by B. Mehta), one by checking the partial-fraction decomposition. Here is another, via the Beta function.

Consider more generally $$ I(a) := \int_0^\infty e^{-nx} \left(\frac{e^{ax}-e^{-ax}}{2a}\right)^{2k} dx $$ for complex $a$ for which the integral converges, that is, with $|{\rm Re}(2ka)| < n$. Then Mr. Bean's integral $I$ is our $I(i)$. Expanding $(e^{ax}-e^{ax})^k$ and integrating termwise we see that $I(a)$ is a linear combination of $1/(n+ja)$ over $j=k,k-2,k-4,\ldots,-k$; in particular, it is a rational function of $a$ and $n$. Hence it is enough to evaluate $I(a)$ for real $a$. But then the change of variables $u = e^{-2ax}$ gives $$ I(a) = \frac1{(2a)^{2k+1}} \int_0^1 u^{n/2a} (u^{-1/2} - u^{1/2})^{2k} \, \frac{du}{u}, $$ which is $$ \frac1{(2a)^{2k+1}} \int_0^1 u^{\frac{n}{2a}-k-1}_{\phantom0} (1-u)^{2k} \, du = \frac1{(2a)^{2k+1}} B(2k+1, \frac{n}{2a} - k). $$ This, in turn, is $$ \frac{(2k)!}{(2a)^{2k+1}} \frac{\Gamma(\frac{n}{2a}-k)}{\Gamma(\frac{n}{2a}+k+1)} = \frac{(2k)!}{(2a)^{2k+1} \prod_{j=-k}^k (\frac{n}{2a}+j)}. = \frac{(2k)!}{\prod_{j=-k}^k (n+2aj)}. $$ Grouping together the $j$ and $-j$ terms, we deduce that $$ I(a) = \frac{(2k)!}{n\prod_{j=1}^k (n^2-(2aj)^2)}. $$ The desired result follows on taking $a=i$. QED

This integral appears as 3.895 #1 on page 478 of Gradshteyn and Ryzhik (with $n,k$ called $\beta, m$), attributed to FI [= Fikhtengolts, Course in differential and integral calculus (1947-1949] Vol. II, 615, and WA [= Watson, A Treatise on the Theory of Bessel Functions, 2nd ed. (1944)] 620a. The analogous formula with odd exponent $$ \int_0^\infty e^{-nx} \sin^{2k+1} x \, dx = \frac{(2k+1)!}{\prod_{j=0}^k (n^2+(2j+1)^2)} $$ appears as 3.895 #4 on the same page, with the exact same attributions; it can be obtained in the same way, or using either of Olivier Oloa's techniques. Even non-integral exponents can be accommodated, though one must be careful to choose the right branches of powers of $\sin x$.

Noam D. Elkies
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2

Following Olivier's idea in the comments, let $I_k = \int_0^\infty e^{-nx} \sin^{2k}\! x \ dx$ for $k\geq1$.

$$\begin{align} I_k &= \int_0^\infty e^{-nx} \sin^{2k}\! x \ dx \\ &=\left[-\frac{1}{n}e^{-n x}\sin^{2k}\! x\right]_0^\infty-\int_0^\infty -\frac{2k}{n}e^{-nx} \sin^{2k-1}\! x \cos x\ dx \\ &=\frac{2k}{n} \int_0^\infty e^{-n x}\sin^{2k-1}\!x\cos x\ dx \\ &=\frac{2k}{n}\left(\left[-\frac{1}{n}e^{-n x}\sin^{2k-1}\!x\cos x\right]_0^\infty-\int_0^\infty -\frac{1}{n}e^{-n x}\frac{d}{dx}(\sin^{2k-1}\!x\cos x)\ dx \right) \\ &=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x \cos^2 x-\sin^{2k}\!x\right)\ dx \\ &=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x (1-\sin^2 x)-\sin^{2k}\!x\right)\ dx \\ &=\frac{2k}{n^2}\int_0^\infty e^{-n x}\left((2k-1)\sin^{2k-2}\!x-2k\sin^{2k}\!x\right)\ dx \\ &=\frac{2k\times(2k-1)}{n^2}I_{k-1}-\frac{2k \times2k}{n^2}I_{k}\\ n^2 I_k &= 2k(2k-1)I_{k-1}-4k^2I_k \\ (n^2+4k^2)I_k&=2k(2k-1)I_{k-1} \\ I_k &= \frac{2k(2k-1)}{n^2+4k^2} I_{k-1} \end{align}$$

We also need to check $I_0$: $$\begin{align}I_0 &= \int_0^\infty e^{-nx}dx \\ &= \left[-\frac{1}{n}e^{-nx}\right]_0^\infty \\ &= \frac{1}{n}\end{align}$$

So, $I_1 = \frac{2\times1}{n^2+4} \frac{1}{n} = \frac{2!}{n^2+4} \frac{1}{n}$, similarly $I_2 = \frac{4\times3}{n^2+4\times2^2} \frac{2!}{n^2+4} \frac{1}{n} = \frac{4!}{(n^2+4\times2^2)(n^2+4)} \frac{1}{n}$, and so on. You can follow the pattern (or, more formally, use induction) to see that $$I_k = \frac{(2k)!}{n \prod_{j=1}^k (n^2 + 4j^2)}$$

B. Mehta
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