Tricky Integral

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June 19th 2011, 02:30 AM
LHeiner

Tricky Integral

Hi

I'm trying to integrate the following:

$\int_{-1}^{1} e^{-\left | t-x \right |} \ dx \ \ \ \ t \in \mathbb{R}$

can you help me how to split the integral up ?

thx
June 19th 2011, 03:13 AM
Prove It

Re: Tricky Integral

Quote:

Originally Posted by LHeiner

Hi

I'm trying to integrate the following:

$\int_{-1}^{1} e^{-\left | t-x \right |} \ dx \ \ \ \ t \in \mathbb{R}$

can you help me how to split the integral up ?

thx

$\displaystyle |t - x| = \begin{cases}t - x \textrm{ if }t - x \geq 0 \implies x \leq t \\ x - t \textrm{ if }t - x < 0 \implies x > t\end{cases}$

So

$\displaystyle \begin{align*} \int_{-1}^1{e^{-|t - x|}\,dx} &= \int_{-1}^1{e^{-(t - x)}\,dt} \textrm{ if }x \leq t \\ &= \int_{-1}^1{e^{x - t}\,dt}\end{align*}$

or

$\displaystyle \begin{align*} \int_{-1}^1{e^{-|t - x|}\,dx} &= \int_{-1}^1{e^{-(x - t)}\,dt} \textrm{ if }x > t\\ &= \int_{-1}^1{e^{t - x}\,dx}\end{align*}$
June 19th 2011, 03:27 AM
Moo

Re: Tricky Integral

Quote:

Originally Posted by Prove It

$\displaystyle |t - x| = \begin{cases}t - x \textrm{ if }t - x \geq 0 \implies x \leq t \\ x - t \textrm{ if }t - x < 0 \implies x > t\end{cases}$

So

$\displaystyle \begin{align*} \int_{-1}^1{e^{-|t - x|}\,dx} &= \int_{-1}^1{e^{-(t - x)}\,dt} \textrm{ if }x \leq t \\ &= \int_{-1}^1{e^{x - t}\,dt}\end{align*}$

or

$\displaystyle \begin{align*} \int_{-1}^1{e^{-|t - x|}\,dx} &= \int_{-1}^1{e^{-(x - t)}\,dt} \textrm{ if }x > t\\ &= \int_{-1}^1{e^{t - x}\,dx}\end{align*}$

You can't write that since you're making a condition on x depending on t, which is a silent variable (in the integral)...
June 19th 2011, 03:37 AM
Prove It

Re: Tricky Integral

Quote:

Originally Posted by Moo

You can't write that since you're making a condition on x depending on t, which is a silent variable (in the integral)...

We are told $\displaystyle t\in \mathbf{R}$ , so it can be treated like a constant.
June 19th 2011, 03:38 AM
Moo

Re: Tricky Integral

Quote:

Originally Posted by LHeiner

Hi

I'm trying to integrate the following:

$\int_{-1}^{1} e^{-\left | t-x \right |} \ dx \ \ \ \ t \in \mathbb{R}$

can you help me how to split the integral up ?

thx

$\begin{aligned} \int_{-1}^1 e^{-|t-x|} ~dx &=\int_{-1}^1 e^{-|t-x|} \bold{1}_{\{x\geq t\}} ~dx+\int_{-1}^1 e^{-|t-x|} \bold{1}_{\{x\leq t\}} ~dx \end{aligned}$

For the first integral, we can write
$\begin{aligned}\int_{-1}^1 e^{-|t-x|} \bold{1}_{\{x\geq t\}} ~dx &=\int_{-1}^1 e^{-|t-x|} \bold{1}_{\{x\geq t\}} \bold{1}_{\{t\leq 1\}} ~dx+\int_{-1}^1 e^{-|t-x|} \bold{1}_{\{x\geq t\}} \bold{1}_{\{t\geq 1\}} ~dx \\ &=\int_{\max(-1,t)}^1 e^{t-x} ~dx \bold{1}_{\{t\leq 1\}} + 0\cdot \bold{1}_{\{t\geq 1\}} \\ &=\begin{cases}\int_{\max(-1,t)}^1 e^{t-x} ~dx &\text{ if } t\leq 1 \\ 0&\text{ if } t\geq 1\end{cases}\end{array}$

Same thing for the second integral, except that we'll consider $t\geq -1$ and $t\leq -1$
June 19th 2011, 03:40 AM
Moo

Re: Tricky Integral

Quote:

Originally Posted by Prove It

We are told $\displaystyle t\in \mathbf{R}$ , so it can be treated like a constant.

Sure it's a constant, but hmmm how to explain it... Have a look at the indicator functions in my post above.
Or try to think about it this way :
You're defining an integral with respect to x. And then you're stating a condition "exterior" to the integral and involving x. That makes no sense at all...
June 19th 2011, 03:54 AM
Moo

Re: Tricky Integral

Quote:

Originally Posted by Moo

$\begin{aligned} \int_{-1}^1 e^{-|t-x|} ~dx &=\int_{-1}^1 e^{-|t-x|} \bold{1}_{\{x\geq t\}} ~dx+\int_{-1}^1 e^{-|t-x|} \bold{1}_{\{x\leq t\}} ~dx \end{aligned}$

For the first integral, we can write
$\begin{aligned}\int_{-1}^1 e^{-|t-x|} \bold{1}_{\{x\geq t\}} ~dx &=\int_{-1}^1 e^{-|t-x|} \bold{1}_{\{x\geq t\}} \bold{1}_{\{t\leq 1\}} ~dx+\int_{-1}^1 e^{-|t-x|} \bold{1}_{\{x\geq t\}} \bold{1}_{\{t\geq 1\}} ~dx \\ &=\int_{\max(-1,t)}^1 e^{t-x} ~dx \bold{1}_{\{t\leq 1\}} + 0\cdot \bold{1}_{\{t\geq 1\}} \\ &=\begin{cases}\int_{\max(-1,t)}^1 e^{t-x} ~dx &\text{ if } t\leq 1 \\ 0&\text{ if } t\geq 1\end{cases}\end{array}$

Same thing for the second integral, except that we'll consider $t\geq -1$ and $t\leq -1$

The end of it...

We get that $\int_{-1}^1 e^{-|t-x|} \bold{1}_{\{x\leq t\}} ~dx=\begin{cases}\int_{-1}^{\min(1,t)} e^{-t+x} ~dx & \text{ if } t\geq -1 \\ 0 & \text{ if } t\leq -1 \end{cases}$

So in the end,

$\int_{-1}^1 e^{-|t-x|} ~dx=\begin{cases} \int_{\max(-1,t)}^1 e^{t-x} ~dx &\text{ if } t\leq -1 \\ \int_{-1}^{\min(1,t)} e^{-t+x} ~dx & \text{ if } t\geq 1 \\ \int_{\max(-1,t)}^1 e^{t-x} ~dx+\int_{-1}^{\min(1,t)} e^{-t+x} ~dx &\text{ if } t\in(-1,1)\end{cases}$