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Evaluate the following definite integrals: (a) int_{0}^{+infty} e^{-st}(t^3 + sin 2t) dt (b) ...

Question:

Evaluate the following definite integrals:

(a) {eq}\int_{0}^{+\infty} e^{-st}(t^3 + \sin 2t)\; dt {/eq}

(b) {eq}\int_{0}^{+\infty} e^{-(s^3 + 4)t} t^2 \;dt {/eq}

Definite Integral

An integral {eq}I {/eq} of the form {eq}I=\int_{x=a}^b f(x)~dx {/eq} is called a definite integral. The values of {eq}a {/eq} are often finite values, that is, numbers but it is possible that {eq}a=-\infty {/eq}. Likewise, The values of {eq}b {/eq} are often finite values, that is, numbers but it is possible that {eq}b=\infty {/eq}. Consequently, we have the four possibilities {eq}I=\int_{x=a}^b f(x)~dx {/eq}, {eq}I=\int_{x=-\infty}^b f(x)~dx {/eq}, {eq}I=\int_{x=a}^\infty f(x)~dx {/eq} and {eq}I=\int_{x=-\infty}^\infty f(x)~dx {/eq} where {eq}a {/eq} and {eq}b {/eq} are finite values, that is, numbers. The definite integrals in the question are of the third type with {eq}a=0 {/eq}. There are several techniques for evaluating a definite integral which include substitution, partial fractions, integration by parts, using a table of integrals, and so on. Sometimes a well known formula for the integral can be used. These ideas are illustrated in the answer that follows.

Answer and Explanation:

The question is restated with slightly different notation. Evaluate the following definite integrals

(a) {eq}~~~I_a=\int_{t=0}^\infty e^{-st}(t^3+\sin(2t))~dt {/eq}

(b) {eq}~~~I_b=\int_{t=0}^\infty e^{-(s^3+4)t}~t^2~dt {/eq}.

Consider part (b). Make the substitution

{eq}\begin{eqnarray*} u &=& (s^3+4)t \\ u^2 &=& (s^3+4)^2t^2 \\ t^2 &=& \frac{u^2}{(s^3+4)^2} \\ \frac{du}{dt} &=& s^3+4 \\ du &=& (s^3+4)dt \\ dt &=& \frac{du}{s^3+4}. \end{eqnarray*} {/eq}

If {eq}t=0 {/eq} then {eq}u=0 {/eq}. If {eq}t=\infty {/eq} then {eq}u=\infty {/eq}. Consequently

{eq}\begin{eqnarray*} I_b &=& \int_{t=0}^\infty e^{-(s^3+4)t}t^2~dt \\ &=& \int_{u=0}^\infty e^{-u}\cdot\frac{u^2}{(s^3+4)^2}\cdot\frac{du}{s^3+4} \\ &=& \frac{1}{(s^3+4)^3}\int_{u=0}^\infty e^{-u}u^2~du \\ &=& \frac{1}{(s^3+4)^3}\int_{u=0}^\infty u^2e^{-u}~du \\ &=& \frac{1}{(s^3+4)^3}\int_{u=0}^\infty u^{3-1}e^{-u}~du \\ &=& \frac{1}{(s^3+4)^3}\cdot\Gamma(3) \\ &=& \frac{1}{(s^3+4)^3}\cdot2! \\ &=& \frac{2}{(s^3+4)^3} \end{eqnarray*} {/eq}

where {eq}\Gamma {/eq} is the gamma function.

Consider part (a). Write the integral as

{eq}\begin{eqnarray*} I_a &=& \int_{t=0}^\infty e^{-st}(t^3+\sin(2t))~dt \\ &=& \int_{t=0}^\infty e^{-st}t^3~dt+\int_{t=0}^\infty e^{-st}\sin(2t)~dt \\ &=& I_1+I_2 \end{eqnarray*} {/eq}

where

{eq}\begin{eqnarray*} I_1 &=& \int_{t=0}^\infty e^{-st}t^3~dt \\ I_2 &=& \int_{t=0}^\infty e^{-st}\sin(2t)~dt. \end{eqnarray*} {/eq}

Consider {eq}I_1 {/eq}. Make the substitution

{eq}\begin{eqnarray*} u &=& st \\ u^3 &=& s^3t^3 \\ t^3 &=& \frac{u^3}{s^3} \\ \frac{du}{dt} &=& s \\ du &=& s~dt \\ dt &=& \frac{du}{s}. \end{eqnarray*} {/eq}

If {eq}t=0 {/eq} then {eq}u=0 {/eq}. If {eq}t=\infty {/eq} then {eq}u=\infty {/eq}. Consequently

{eq}\begin{eqnarray*} I_1 &=& \int_{t=0}^\infty e^{-st}t^3~dt \\ &=& \int_{u=0}^\infty e^{-u}\cdot\frac{u^3}{s^3}\cdot\frac{du}{s} \\ &=& \frac{1}{s^4}\int_{u=0}^\infty e^{-u}u^3~du \\ &=& \frac{1}{s^4}\int_{u=0}^\infty u^3e^{-u}~du \\ &=& \frac{1}{s^4}\int_{u=0}^\infty u^{4-1}e^{-u}~du \\ &=& \frac{1}{s^4}\cdot\Gamma(4) \\ &=& \frac{1}{s^4}\cdot3! \\ &=& \frac{6}{s^4}. \end{eqnarray*} {/eq}

Consider {eq}I_2 {/eq}. Either use a table of integrals or apply integration by parts twice. Consider integration by parts. The limits of integration will be put in the final answer. Let

{eq}\begin{eqnarray*} u &=& e^{-st} \\ dv &=& \sin(2t)~dt \\ du &=& -se^{-st}~dt \\ v &=& -\frac{1}{2}\cos(2t). \end{eqnarray*} {/eq}

Then

{eq}\begin{eqnarray*} I_2 &=& uv-\int v~du \\ &=& -\frac{1}{2}e^{-st}\cos(2t)-\int\left(-\frac{1}{2}\cos(2t)\right)(-se^{-st}~dt) \\ &=& -\frac{1}{2}e^{-st}\cos(2t)-\frac{s}{2}\int\cos(2t)e^{-st}~dt. \end{eqnarray*} {/eq}

Apply integration by parts again as follows

{eq}\begin{eqnarray*} u &=& e^{-st} \\ dv &=& \cos(2t)~dt \\ du &=& -se^{-st}~dt \\ v &=& \frac{1}{2}\sin(2t). \end{eqnarray*} {/eq}

Then

{eq}\begin{eqnarray*} \int\cos(2t)e^{-st}~dt &=& uv-\int v~du \\ &=& \frac{1}{2}e^{-st}\sin(2t)-\int\left(\frac{1}{2}\sin (2t)\right)(-se^{-st}~dt) \\ &=& \frac{1}{2}e^{-st}\sin(2t)+\frac{s}{2}\int\sin(2t)e^{-st}~dt \\ &=& \frac{1}{2}e^{-st}\sin(2t)+\frac{s}{2}~I_2. \end{eqnarray*} {/eq}

Combine this with the first application of integration by parts to get

{eq}\begin{eqnarray*} I_2 &=& -\frac{1}{2}e^{-st}\cos(2t)-\frac{s}{2}\left(\frac{1}{2}e^{-st}\sin(2t)+\frac{s}{2}~I_2\right) \\ I_2 &=& -\frac{1}{2}e^{-st}\cos(2t)-\frac{s}{4}e^{-st}\sin(2t)-\frac{s^2}{4}~I_2 \\ I_2+\frac{s^2}{4}~I_2 &=& -\frac{1}{2}e^{-st}\cos(2t)-\frac{s}{4}e^{-st}\sin(2t) \\ \left(\frac{4+s^2}{4}\right)I_2 &=& -\frac{e^{-st}(2\cos(2t)+s\sin(2t))}{4} \\ (4+s^2)I_2 &=& -e^{-st}(2\cos(2t)+s\sin(2t)) \\ I_2 &=& -\frac{e^{-st}(2\cos(2t)+s\sin(2t))}{4+s^2}. \end{eqnarray*} {/eq}

Now include the limits of integration to get

{eq}\begin{eqnarray*} I_2 &=& \left.-\frac{e^{-st}(2\cos(2t)+s\sin(2t))}{4+s^2}\right|_{t=0}^\infty \\ &=& 0-\left(-\frac{2}{4+s^2}\right) \\ &=& \frac{2}{4+s^2}. \end{eqnarray*} {/eq}

Therefore

{eq}\begin{eqnarray*} I_a &=& I_1+I_2 \\ &=& \frac{6}{s^4}+\frac{2}{4+s^2} \\ &=& \frac{2(3(4+s^2)+s^4)}{s^4(4+s^2)} \\ &=& \frac{2(12+3s^2+s^4)}{s^4(4+s^2)} \\ &=& \frac{24+6s^2+2s^4}{4s^4+s^6}. \end{eqnarray*} {/eq}


Learn more about this topic:

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Evaluating Definite Integrals Using the Fundamental Theorem

from AP Calculus AB: Exam Prep

Chapter 16 / Lesson 2
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