# Integrate ({e^x}logx)

1. Jan 2, 2009

### FedEx

$$\int e^xlog(x)$$

I have solved this sum using integration by parts. The answer which i get is

$$e^xlogx - logx - \sum^{\infty}_{i=1}\frac{x^i}{(i)(i!)}$$

But i also used the series expansion of e^x. Is there any other way of doing this sum?

I have almost tried out every single way of doing this sum by parts. So better think about substitution or any other possible method other then by parts.

Last edited: Jan 2, 2009
2. Jan 2, 2009

### MathematicalPhysicist

substitute e^x=u, you get e^xdx=du, dx=du/u, you can fill the blanks by your own.

3. Jan 2, 2009

### HallsofIvy

Staff Emeritus
If you know that $\int ln x dx= x ln x - x+ C$, which you can do by integration by parts, then it is easy to take u= ex, dv= ln x and use integration by parts.

4. Jan 2, 2009

### FedEx

But on doing so we will get

$$log(log u) du$$ to integrate. And i did integrate that but i had to use series expansion of e^x.

5. Jan 2, 2009

### FedEx

I am sorry but i am not able to get you.

Integration of log x is simple. Here what do you mean by dv? Me being in India have a slight of different version of notations. Do you mean that we take log x as the second function? I have tried that but i get

$$xe^xlogx$$ to integrate. Which i am unable to do

6. Jan 2, 2009

### D H

Staff Emeritus
That's probably what the OP did, with one error.

How does this help?

That looks good except for one term. Could you show your derivation?

This integral does not have a solution in the elementary functions. In other words, the only way to express the integral is to use either series as you have done or to use some special function. If you choose the appropriate values for u and dv, you will get a rather widely used special function in one step.

Note: I'm not saying you should use that particular special function. Your instructor may well want the series representation.

7. Jan 2, 2009

### FedEx

I am not able to get this.

Take

$$x= log\alpha$$

By parts i get

$$\alpha log(log\alpha) - \int\frac{d\alpha}{log\alpha}$$

let $$log\alpha = \beta$$

we get

$$\alpha log(log\alpha) - \int\frac{e^{\beta}d\beta}{\beta}$$

Then i apply series and integrate

$$\alpha log(log\alpha) - \frac{1}{{\beta}^2} - \beta - \frac{{\beta}^2}{(2)(2!)} - \frac{{\beta}^3}{(3)(3!)} .........$$

Now bringing it back in x
$$e^xlogx -logx - \sum^{\infty}_{i=1}\frac{x^i}{(i)(i!)}$$

Last edited: Jan 2, 2009
8. Jan 2, 2009

### D H

Staff Emeritus
Wow. That was a hard way to get to this result. Look at what you have done. $\beta$ is just $x$. You assigned $x= \log\alpha$ and $\log\alpha = \beta$. By transitivity, $\beta=x$.

Since you arrived at this result the hard way, here's an easy way: Integrate by parts directly.

Let

\aligned u&=\log x &\quad dv &=e^x dx \\ du &= \frac {dx}x & \quad v &= e^x \endaligned

Thus

$$\int e^x\log x dx = e^x\log x - \int \frac{e^x}{x} dx$$

Too fast! Where does the $1/beta^2$ come from?

9. Jan 2, 2009

### FedEx

Ya.I saw it at last while putting back the values in terms of x.

$$\frac{e^{\beta}}{\beta} = \frac{{1} + \beta + \frac{{\beta}^2}{2!} + \frac{{\beta}^3}{3!} + ......}{\beta}$$

Then i integrated it.

10. Jan 2, 2009

### D H

Staff Emeritus
What is the integral of 1/beta? (Hint: It is not 1/beta2).

11. Jan 2, 2009

### FedEx

Oh my god. How can this happen. I am so sorry. I am a fool. Its

$$log{\beta}$$

I am so sorry. I a ashamed of myself. May be its because i am ill at the moment. However illness has nothing to do with it this is just because of my mere lack of concentration

12. Jan 2, 2009

### D H

Staff Emeritus
Naw. You just made a dumb mistake. We all make dumb mistakes. In general, admit it (eat some crow) and move on. Crow can be a tasty dish at times.

Now that you have the answer in series form, there is a way to express it without a series using the "exponential integral":

$$\textrm{Ei}(x) \equiv \int_{-\infty}^x \frac {e^t}t dt$$

13. Jan 2, 2009

### FedEx

Thanks for the consolation.

So you mean that the sigma and the log x term can be replaced by the exponential integral, isn't it?

14. Jan 2, 2009

### D H

Staff Emeritus
Yes, but if you haven't covered special functions yet your instructor probably wants to see the series representation.

15. Jan 3, 2009

### FedEx

Thanks a lot D H. You really helped me with the sum