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November 21

Integration by parts involving e and sin

How do you solve an integral that includes a function including e multiplied by a function including sin. I keep having to integrate by parts over and over again and it doesn't get me anywhere as e just differentiates and integrates to itself. 194.66.246.41 (talk) 10:37, 21 November 2014 (UTC)[reply]

Often, after a few iterations you get back the integral you started with. Then you treat this integral as a variable and solve the equation for it. At other times, the best solution you can find is an infinite series. -- Meni Rosenfeld (talk) 12:34, 21 November 2014 (UTC)[reply]

Have you tried expressing the sin as the sum of complex exponentials? See Euler's formula and the section relation to trigonometry. p,s, you can see the integral of an exponential multiplied by sin at List of integrals of exponential functions, but then you wouldn't have the fun of doing it yourself would you ;-) Dmcq (talk) 13:06, 21 November 2014 (UTC)[reply]

I wouldn't normally do it but one of the yahoo answers pages really looks good on this. https://answers.yahoo.com/question/index?qid=20080120225443AAVhiAm Naraht (talk) 16:05, 21 November 2014 (UTC)[reply]

To elaborate on Meni Rosenfeld's method:

\int e^{x}\sin(x)dx=-e^{x}\cos(x)+\int e^{x}\cos(x)dx=-e^{x}\cos(x)+e^{x}\sin(x)-\int e^{x}\sin(x)dx

Then add

\int e^{x}\sin(x)dx

to the left side:

2\int e^{x}\sin(x)dx=-e^{x}\cos(x)+e^{x}\sin(x)

\int e^{x}\sin(x)dx={\frac {-e^{x}\cos(x)+e^{x}\sin(x)}{2}}+C

This is often taught in math textbooks as "solving for the unknown integral" and is applicable to any integral of this form.--Jasper Deng (talk) 18:41, 21 November 2014 (UTC)[reply]

Abs Value Inequality Question

I'm trying to follow a proof that takes a jump I can't see. It says "it follows from |f(x) - L| < 1 that |f(x)| < |L| + 1." I can work out -1 < f(x) - L < 1, then get f(x) < L + 1 from the middle and right terms, then reason since L is always less than or equal to |L|, that f(x) < |L| + 1. But how from here can I get to saying the absolute value of f(x) is less? Peter Michner (talk) 16:59, 21 November 2014 (UTC)[reply]

One way is to use the fact that |f(x) - L| = |L - f(x)|, giving you |L - f(x)| < 1, then repeat what you did before. Another is to take the left and middle terms your -1 < f(x) - L < 1 and multiply both of those side by -1, changing the sign of all terms and reversing the inequality. I'll give more than a hint if you wish, but I thought you might enjoy working it out here yourself. -- ToE 17:40, 21 November 2014 (UTC)[reply]

Alternately, just use the triangle inequality directly. This is often expressed as |x + y| ≤ |x| + |y|, but by substituting x = a - b and y = b, you get |a| ≤ |a - b| + |b| which you can rearrange as |a| - |b| ≤ |a - b|. Apply that to your formula and you get |f(x)| - |L| ≤ |f(x) - L| < 1. Take the left and right terms, |f(x)| - |L| < 1, and add |L| to both sides. -- ToE 17:52, 21 November 2014 (UTC)[reply]

I see that |a| - |b| ≤ |a - b| is part of what is called the reverse triangle inequality. -- ToE 18:04, 21 November 2014 (UTC)[reply]