# 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.

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}