Evaluate the indefinite integral as an infinite series.?
Take the series for e^x : e^x = 1 + x + x^2/ 2! + x^3 /3! + x^4 /4! + ... Now subtract 1 and divide each term in the series by x to get: (e^x -1)/x = 1 + x/ 2! + x^2/ 3! + x^3 /4! + ... Now integrate term by term (Note: not all infinite series can be integrated term by term; however, power series can always be integrated term by term in the interval of convergence): So the integral will be: x + x^2 /( 2*2!) + x^3 /(3*3!) + x^4 /(4*4!) + .... + C (the constant of integration) And that can be written as Σ n = 1 to infinity of: x^n /(n*n!) + C.
Express the definite integral as an infinite series ∫0 to 1 cos((5/6)x^4)dx?
cos t = Σ ( -1) ^n x^(2n) / ( 2n)!...thus cos (5x^4 / 6 ) = Σ (-1)^n [ (5x^4/6)^2n / (2n)! ]...integrate.... evaluate at x = 1...it is an alternating series...make the 1st term not used to be < 10^(-4)............. 3 terms likely adequate
Evaluate the indefinite integral as an infinite series: integral of (cos(x) - 1)/x dx?
Since cos x = sum(k=0 to infinity) (-1)^k x^(2k)/(2k)!, (cos x - 1)/x = [-1 + sum(k=0 to infinity) (-1)^k x^(2k)/(2k)!] / x = sum(k=1 to infinity) (-1)^k x^(2k-1)/(2k)! Integrating term by term yields int (cos x - 1) dx /x = C + sum(k=1 to infinity) (-1)^k x^(2k)/[(2k)(2k)!]. I hope that helps!
How do I evaluate an indefinite integral as an infinite series?
Evaluate the indefinite itnegral as an infinite series: 3∫(e^x - 1)/5(x) dx The sum must begin from n = 1. I have no idea how to do this. If somebody could please explain to me how to do it, I would really appreciate it. Do I just differentiate it for a few terms, and plug in 0 into each derivative like I would with a regular function that is not an integral? Thank you.
Express the integral from 0 to x of (1-cos x)/x as an infinite series?
Since cos t = Σ(n=0 to ∞) (-1)^n t^(2n)/(2n)! = 1 + Σ(n=1 to ∞) (-1)^n t^(2n)/(2n)!: (1 - cos t)/t = {1 - [1 + Σ(n=1 to ∞) (-1)^n t^(2n)/(2n)!]} / t ................= Σ(n=1 to ∞) (-1)^(n+1) t^(2n-1)/(2n)!. Therefore, ∫(0 to x) (1 - cos t) dt/t = Σ(n=1 to ∞) (-1)^(n+1) x^(2n)/[(2n) (2n)!]. I hope this helps!
Evaluate integral( e^x/x )dx as an infinite series?
e^x = 1 + x + x^2/2! + x^3/3! + ... e^x/x = 1/x + 1 + x/2! + x^2/3! + ... ∫ (e^x/x) dx = ∫ (1/x + 1 + x/2! + x^2/3! + ...) dx = ln x + x + x^2/(2 • 2!) + x^3/(3 • 3!) + ... + C
Evaluate the integral e^x / x as an infinite series?
∫ [e^x / x ] dx ∫ (e^ x) [(f(x) + f'(x)] dx = (e^x)f((x) + constant can easily proved by parts for ∫ {[e^x / x ] dx = ∫ (e^x) [ f(x) + f'(x)] - [f'(x) + f"(x)] + [f"(x)- f"'(x)] - [f"'(x) - f""x} ------------}dx = ∫ {[e^x) [(1/x) + (-1)/ x²] - [(-1)/ x² + (-1)(-2)/ x³] + [(-1)(-2)/ x³ + (-1)(-2)(-3)/ x4] ------------ } dx = ∫ [e^x) [(1/x) + (-1)/ x²] - ∫e^x) [(-1)/ x² + (-1)(-2)/ x³]dx + ∫ e^x)[(-1)(-2)/ x³ + (-1)(-2)(-3)/ x4]dx + .......∫ = [e^x][ (1/x) - (-1/x²) + (-1)(-2)/ x³ - (-1)(-2)(-3)/ x4 + -------------∞ = e^x [ 1/x + 1/x² + (1*2)/x³ +(1*2*3)/ x4 + (1*2*3*4)/x^5 ----------+ (n-1)! / x^n] as n --- ∞ = (e^x) ∑[(n-1)! / x^n] [1, ∞ ]
Express the integral of (e^x - 1)/x as an infinite series.?
Starting with e^x = 1 + x + x^2/2! + x^3/3! + ... + x^n/n! + ... ==> e^x - 1 = x + x^2/2! + x^3/3! + ... + x^n/n! + ... ==> (e^x - 1)/x = 1 + x/2! + x^2/3! + ... + x^(n-1)/n! + ... = Σ(n=1 to ∞) x^(n-1)/n!. I hope this helps!
How do I solve integral from 0 to infinity cos(x^2)?
Are you aware of the Gaussian integral? From the wiki,[math]\int_{-\infty}^{\infty}e^{-x^2}dx = \sqrt{\pi}[/math]The wiki provides multiple proofs for the same, so I won't replicate it here. We will be using this to get the value of the desired definite integral.Claim 1: [math]\int_{0}^{\infty}e^{-x^2}dx = \frac{\sqrt{\pi}}{2}[/math]Proof: Since [math]e^{-x^2}[/math] is an even function.Claim 2: [math]\int_{0}^{\infty}e^{ix^2}dx = \frac{e^{i\pi / 4}}{2}\sqrt{\pi}[/math]Proof: Let [math]I = \int_{0}^{\infty}e^{ix^2}dx[/math]. Substituting [math]x = e^{i\pi / 4}y[/math], we get,[math]I = \int_{0}^{\infty}e^{i\pi / 4}e^{-y^2}dy = \frac{e^{i\pi / 4}}{2}\sqrt{\pi}[/math]Note that the proof of Claim 2 is not entirely rigorous. Ideally, we have to argue that the complex integral is indeed the same as integrating over the positive real axis. I am somewhat glossing over it at the moment, but the argument is standard, so if you have experience in complex integration, you should be able to fill it in.Claim 3: [math]\int_{0}^{\infty}\cos x^2 dx = \sqrt{\frac{\pi}{8}}[/math]Proof: From Claim 2,[math]\int_{0}^{\infty}e^{ix^2}dx = \frac{e^{i\pi / 4}}{2}\sqrt{\pi}[/math][math]\int_{0}^{\infty}\cos x^2 dx + i \int_{0}^{\infty}\sin x^2 dx = \frac{\sqrt{\pi}}{2}\left(\frac{1}{\sqrt{2}} + i\frac{1}{\sqrt{2}}\right)[/math]Comparing the real parts, gives us the required answer.
Pls integrate (e^-x)(cos(x))dx from 0 to infinity?
I = ∫ e^-x cosx dx = e^-x ∫ cosx dx - ∫ [d/dx(e^-x) ∫ cosx dx]dx = e^-x sinx + ∫ e^-x sinx dx = e^-x sinx + [e^-x ∫ sinx dx - ∫ [d/dx(e^-x) ∫ sinx dx] dx] = e^-x sinx - e^-x cosx - ∫ e^-x cosx dx + 2c => 2I = e^-x (sinx - cosx) + 2c => I = (1/2) e(-x)(sinx - cosx) + c Plugging limits x = 0 to x = infinity I = (1/2) (0 - 0 + 1) => I = 1/2.