$\displaystyle \int e^{-\lambda(t-y)} dy= \frac{1}{\lambda}e^{-\lambda(t-y)} +C= F(y)$

By the fundamental theorem of calculus:

$\displaystyle \int_{0}^{x}e^{-\lambda(t-y)} dy=F(x)-F(0)=\frac{1}{\lambda}(e^{-\lambda{(t-x)}})-\frac{1}{\lambda}(e^{-\lambda{(t-0)}})$

$\displaystyle \int_{0}^{x}e^{-\lambda(t-y)} dy=\frac{1}{\lambda}[e^{-\lambda({t-x})}-e^{-\lambda t}]$

First compute the indefinite integral. Remember that

$\displaystyle \int e^x=e^x$

Evaluate the integral from the limits given

$\displaystyle e^x|\frac{\infty }{0} =e^\infty-e^0$

$\displaystyle e^\infty$ is always $\displaystyle \infty$ and any number to the power of 0 is always 1.

therefore: $\displaystyle \infty-1=\infty$

This intergral is improper because of the upper limit at infinity. To handle this, we use the theorem that if an integral 

$\displaystyle \int_a^{\infty}f(x)dx$ 

exists, then it is equal to the value of 

$\displaystyle \lim_{b\rightarrow\infty}\int_a^bf(x)dx$.

Applying this to our given integral, we get:

$\displaystyle \int_1^{\infty}\frac{1}{x^n}dx=\lim_{b\rightarrow\infty}\int_1^b\frac{1}{x^n}dx=\lim_{b\rightarrow\infty}\int_1^bx^{-n}dx$

Using the power rule for integration, which is valid since $\displaystyle n\neq1$,we obtain:

$\displaystyle \lim_{b\rightarrow\infty}\frac{x^{-n+1}}{-n+1}\biggr|_1^b=\lim_{b\rightarrow\infty}\biggr(\frac{b^{-n+1}}{-n+1}-\frac{1^{-n+1}}{-n+1}\biggr)$

$\displaystyle =\lim_{b\rightarrow\infty}\biggr(\frac{b^{-n+1}}{-n+1}\biggr)-\frac{1}{-n+1}$

Here the second term in the limit didn't depend on $\displaystyle b$, so it evaluates to 

$\displaystyle -\frac{1}{-n+1}=\frac{1}{n-1}$.

To evaluate the limit for the first term, we note that from $\displaystyle n>1$, we have that $\displaystyle -n< -1\rightarrow-n+1< 0$.

Therefore the power on $\displaystyle b$ is always less than $\displaystyle 0$ for any $\displaystyle n>1$. We have the theorem that 

$\displaystyle \lim_{b\rightarrow\infty}b^x=0 \iff x< 0$.

Therefore the first term in the limit as $\displaystyle b\rightarrow\infty$ vanishes, and we are left with the result 

$\displaystyle \int_1^\infty\frac{1}{x^n}dx=\frac{1}{n-1}$.

The integral we are given is improper due to the upper bound being infinity. To integrate, we must rewrite the integral with a limit for the upper bound:

$\displaystyle \frac{1}{3}\lim_{a\rightarrow \infty^+}\int_{1}^{a} \frac{dx}{x}$

Note that we moved the constant outside of the integral and the limit. 

Now, we can integrate, remembering to keep the limit in front of every step in the process:

$\displaystyle \frac{1}{3}\lim_{a\rightarrow \infty^+}\int_{1}^{a} \frac{dx}{x}=\frac{1}{3}\lim_{a\rightarrow \infty^+} \ln \left | x\right |\Big|^{0}_{a}$

The integral was performed using the following rule:

$\displaystyle \int\ \frac{dx}{x}=\ln\left | x\right |+C$

Now, we can definitely integrate by plugging in the upper and lower bounds and subtracting the results:

$\displaystyle \frac{1}{3}\lim_{a\rightarrow \infty^+} \ln(a)-\ln(1)=\infty$

Note that the absolute value goes away because the bounds of the integral are both positive.

The function goes to infinity as the natural log of infinity is infinity.

Substitute a in for infinity and solve the integral:

$\displaystyle \int^{a}_{1/2} \frac{3}{x^4}dx=-\frac{1}{x^3}\left.\begin{matrix} \\ \end{matrix}\right|^a_{1/2}=-\frac{1}{a^3}-(-\frac{1}{(1/2)^3})=8-\frac{1}{a^3}$

Now, take the limit of this expression as a approaches infinity:

$\displaystyle \lim_{a\rightarrow \infty}(8-\frac{1}{a^3})=8$

To evaluate the improper integral (infinity is its upper limit of integration which makes it improper), we must use a limit for the upper bound.

But before we do this, we can use a u-substitution to rewrite the integral as something we can easily integrate:

$\displaystyle u=\ln(x)$, $\displaystyle du=\frac{dx}{x}$

The derivative is found using the identical rule to the derivative.

Now, rewrite the integral, with a limit for the upper bound, and integrate

$\displaystyle \lim_{t\rightarrow \infty}\int_{x=2}^{x=t}\frac{du}{u^3}=\lim_{t\rightarrow \infty} \int_{x=2}^{x=t}-\frac{1}{2u^2}$

The integral was found using the following rule:

$\displaystyle \int x^ndx=\frac{x^{n+1}}{n+1}+C$

Next, we can replace u with our x terms, and perform the definite integration:

$\displaystyle \lim_{t\rightarrow \infty}-\frac{1}{2(\ln x)^2}\left.\begin{matrix} t\\2 \end{matrix}\right|=\lim_{t\rightarrow \infty}-\frac{1}{2(\ln t)^2}+\frac{1}{2 \ln(2)^2}=0+\frac{1}{2\ln(2)^2}$

As t goes to infinity, the denominator of the first term goes to infinity, which means the entire term goes to zero. We are left with

$\displaystyle \frac{1}{2(\ln2)^2}$

To evaluate the improper integral (infinity is its upper bound), we must use a limit on the upper bound:

$\displaystyle \lim_{t\rightarrow \infty^+}\int_{1}^{t}\frac{dx}{4x}$

Note that we are approaching infinity from the right because it is the upper bound of integration. 

We can now integrate (don't forget the limit!):

$\displaystyle \lim_{t\rightarrow \infty^+} \frac{1}{4}\ln\left | x\right |\left.\begin{matrix} t\\1 \end{matrix}\right|$

The following rule was used for integrating:

$\displaystyle \int \frac{dx}{x}=\ln\left | x\right |+C$

Now, the definite part of the integration:

$\displaystyle \lim_{t\rightarrow \infty^+}\frac{1}{4}(\ln(t)-\ln(1))=\infty$

Note that because the bounds of the integral were both positive, we removed the absolute value sign. When t goes to infinity, the natural logarithm function goes to infinity. The natural logarithm of 1 is zero, so we are left with infinity.

This integral is improper because the lower bound creates a zero in the denominator. To integrate, we must use a limit:

$\displaystyle \lim_{t\rightarrow 1^-}\int_{t}^{2}\frac{dx}{x-1}=\lim_{t\rightarrow 1^-} \ln\left | x-1\right |\left.\begin{matrix} 2\\ t \end{matrix}\right|$

The following rule was used:

$\displaystyle \int \frac{dx}{x}=\ln\left | x\right |+C$

Now, the definite part of the integral:

$\displaystyle \lim_{t\rightarrow 1^-} \ln\left | 1\right |-\ln\left | t-1\right |=\infty$

The natural logarithm of 1 equals zero, and as the natural logarithm approaches zero, the function goes to negative infinity. The negative sign in front of the function makes it go to infinity.

We can evaluate this integral using integration by parts

$\displaystyle \int_0^{\infty} te^{-st}dt$. (Start)

$\displaystyle = [-\frac{te^{-st}}{s}]^\infty_0 - \int_0^\infty -\frac{e^{-st}}{s}dt$ (Use integration by parts, with $\displaystyle u = t, dv =e^{-st}dt$)

$\displaystyle = [(\lim_{t \to \infty}-\frac{te^{-st}}{s})-(-\frac{0}{s})]-[\frac{e^{-st}}{s^2}]_0^\infty$

$\displaystyle =[0-0]-[\lim_{t \to \infty} \frac{e^{-st}}{s^2}-\frac{1}{s^2}]$

$\displaystyle =[0]-[0-\frac{1}{s^2}]$

$\displaystyle =\frac{1}{s^2}$.