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Again:

$$\int e^{\alpha x}\cos(\beta x) \space dx = \frac{e^{\alpha x} (\alpha \cos(\beta x)+\beta \sin(\beta x))}{\alpha^2+\beta^2}$$

Also the one for $\sin$:

$$\int e^{\alpha x}\sin(\beta x) \space dx = \frac{e^{\alpha x} (\alpha \sin(\beta x)-\beta \cos(\beta x))}{\alpha^2+\beta^2}$$

Both help avoid integration by parts. I find these two proofs very useful and I'd like to know what their names are.

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    $\begingroup$ You can use \sin and \cos in LaTeX to make your problem more readable. $\endgroup$
    – Tyler
    Commented Feb 1, 2011 at 2:04
  • $\begingroup$ @tyler: ah thanks! I should've thought of that. It's my first time actually writing something complicated. $\endgroup$ Commented Feb 1, 2011 at 2:07
  • $\begingroup$ @Tyler Bailey: Just wondering, how did you edit that post? Don't you need 2000 reputation? $\endgroup$ Commented Feb 1, 2011 at 2:12
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    $\begingroup$ What you have given here is an identity, and can ask for it's name. You haven't actually given any proof... $\endgroup$
    – Aryabhata
    Commented Feb 1, 2011 at 2:26
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    $\begingroup$ I seriously doubt those equalities have a name: they are just indefinite integrals, and there are way too many of those to name them... $\endgroup$ Commented Feb 1, 2011 at 3:52

3 Answers 3

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They are the real and imaginary parts of the following identity: $$ \int{e^{(\alpha+i\beta)x}dx} = \frac{e^{(\alpha+i\beta)x}}{\alpha+i\beta} = \frac{(\alpha - i\beta)e^{(\alpha+i\beta)x}}{\alpha^2 + \beta^2}. $$

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  • $\begingroup$ Does it have a name? $\endgroup$ Commented Feb 1, 2011 at 3:04
  • $\begingroup$ and also: Wait, what? Could you give some explanation on how that relates to my identity? I don't understand how trig becomes something containing $i$ $\endgroup$ Commented Feb 1, 2011 at 3:07
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    $\begingroup$ @JustcallmeDrago Take a look at Euler's formula and you'll see the relationship. $\endgroup$ Commented Feb 1, 2011 at 3:15
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    $\begingroup$ @mjqxxxx Now that is a very clever way to find them! $\endgroup$
    – Pedro
    Commented Apr 29, 2012 at 1:01
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I answered this in another question, but I think it might apply better here:


In general, if you want to find $$ \int e^{ax}\cdot \sin{bx}\cdot dx$$ you can argue as follows:

Note that for any $\alpha$ or $\beta$, you have

$$\eqalign{ & \frac{d}{{dx}}\left( {{e^{\alpha x}}\sin \beta x} \right) = \alpha {e^{\alpha x}}\sin \beta x + \beta {e^{\alpha x}}\cos \beta x \cr & \frac{d}{{dx}}\left( {{e^{\alpha x}}\cos \beta x} \right) = \alpha {e^{\alpha x}}\cos \beta x - \beta {e^{\alpha x}}\sin \beta x \cr} $$

so that any integral of the form

$$ \int e^{\alpha x}\cdot \sin{\beta x}\cdot dx$$

is a linear combination of the former functions. Let's then find $c_1$ and $c_2$ such that

$$\frac{d}{{dx}}\left( {{c_1}{e^{\alpha x}}\sin \beta x + {c_2}{e^{\alpha x}}\cos \beta x} \right) = {e^{\alpha x}}\sin \beta x$$

$${c_1}\alpha {e^{\alpha x}}\sin \beta x + {c_1}\beta {e^{\alpha x}}\cos \beta x + {c_2}\alpha {e^{\alpha x}}\cos \beta x - {c_2}\beta {e^{\alpha x}}\sin \beta x = {e^{\alpha x}}\sin \beta x$$

This means we need

$$\eqalign{ & {c_1}\alpha - {c_2}\beta = 1 \cr & {c_1}\beta + {c_2}\alpha = 0 \cr} $$

This will yield with little work

$$\eqalign{ & {c_1} = \frac{\alpha }{{{\alpha ^2} + {\beta ^2}}} \cr & {c_2} = - \frac{\beta }{{{\alpha ^2} + {\beta ^2}}} \cr} $$

which means that, in general:

$$\int {{e^{\alpha x}}} \cdot\sin \beta x\cdot dx = {e^{\alpha x}}\frac{{\alpha \sin \beta x - \beta \cos \beta x}}{{{\alpha ^2} + {\beta ^2}}} + C$$

Analogously, you will get that

$$\int {{e^{\alpha x}}} \cdot\cos \beta x\cdot dx = {e^{\alpha x}}\frac{{\alpha \cos \beta x + \beta \sin \beta x}}{{{\alpha ^2} + {\beta ^2}}} + C$$


This has no formal name, it is just a formula for finding a primitive.

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HINT

Generally, if we denote $A=\int e^{a x} \cdot \cos (b x) \ dx$ and $B=\int e^{a x} \cdot \sin (b x) \ dx$ and integrate by parts A and B, then we obtain an easy-to-solve system in $A$ and $B$. But this is just another possible approach. However, mjqxxxx's approach is by far a better one.

Chris.

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